Circle

Circle 1 given $x^2+y^2=4$

Circle 2 is given,

$ x^2+y^2-6 x-6 y+14=0 $

Centre $(3,3)$ radius $=2$

Let $P\left(x_1, y_1\right)$ be a point on circle 1 .

$\because$ Equation of chord of contact $A B$ is

$ \begin{aligned} & x x_1+y y_1-3\left(x+x_1\right)-3\left(y+y_1\right)+14=0 \\ & \left(x_1-3\right) x+\left(y_1-3\right) y-3 x_1-3 y_1+14=0 \end{aligned} $

Now, equation of circle passing through $P, A$ and $B$ is

$ \begin{aligned} x^2+y^2-6 x-6 y+14 +\lambda\left[\left(x_1-3\right) x+\left(y_1-3\right) y-3 x_1\right. & \left.-3 y_1+14\right]=0 \end{aligned} $

We have this circle passes through $P\left(x_1, y_1\right)$

$ \begin{array}{r} \\\Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda\left[\left(x_1-3\right) x_1\right. \\ \left.+\left(y_1-3\right) y_1-3 x_1-3 y_1+14\right]=0 \\ \Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda\left[x_1^2+y_1^2\right. \\ \left.-3 x_1-3 y_1-3 x_1-3 y_1+14\right]=x \end{array} $

Put $x_1^2+y_1^2=4$

$ \begin{gathered} \Rightarrow 4-6 x_1-6 y_1+14 \\ +\lambda\left(4-6 x_1-6 y_1+14\right)=0 \\ \Rightarrow 18-6 x_1-6 y_1+14+\lambda\left(18-6 x_1\right. \\ \left.-6 y_1\right)=0 \end{gathered} $

$ \begin{aligned} & \because 1+\lambda=0 \\ & \lambda=-1 \end{aligned} $

$\Rightarrow$ Equation of circle be

$ \begin{aligned} & \left(x^2+y^2-6 x-6 y+14\right) \\ & -\left[\left(x_1-3\right) x+\left(y_1-3\right) y\right. \\ & \left.-3 x_1-3 y_1+14\right]=0 \\ & \Rightarrow x^2+y^2-\left(3+x_1\right) x-\left(3+y_1\right) y \\ & +3 x_1+3 y_1=0 \end{aligned} $

Centre $\left(\frac{3+x_1}{2}, \frac{3+y_1}{2}\right)$

$ h=\frac{3+x_1}{2} \text { and } k=\frac{3+y_1}{2} $

$ \begin{aligned} & x_1=2 h-3 \quad y_1=2 k-3 \\ & x_1^2+y_1^2=4 \\ & (2 h-3)^2+(2 k-3)^2=4 \\ & 4 h^2+4 k^2-12 h-12 k+14=0 \\ & h^2+k^2-3 h-3 k+\frac{7}{2}=0 \end{aligned} $

$\because$ Taking locus of $(h, k)$, we get

$ 2 x^2+2 y^2-6 x-6 y+7=0 $

Let the ends of diameter of the given circle $x^2+y^2=4$ is

$(2 \cos \theta, 2 \sin \theta)$, then other must be ( $-2 \cos \theta,-2 \sin \theta$ )

Then, length of perpendicular from the extremities of diameter to the line $x+y+1=0$ is

$ \begin{aligned} & \frac{2(\cos \theta+\sin \theta)+1}{\sqrt{2}} \\ & \times \frac{-2(\cos \theta+\sin \theta)+1}{\sqrt{2}}=P(\text { say }) \\ & \Rightarrow P=\frac{1-4(\cos \theta+\sin \theta)^2}{2} \\ & =\frac{1-4(1+\sin 2 \theta)}{2} \\ & \Rightarrow P=\frac{-3-4 \sin 2 \theta}{2} \\ & \Rightarrow \quad \frac{d P}{d \theta}=\frac{1}{2}(-4) \cos 2 \theta \cdot 2=0 \\ & \Rightarrow \quad \cos 2 \theta=0 \Rightarrow 2 \theta=\frac{\pi}{2} \\ & \quad \theta=\frac{\pi}{4} \end{aligned} $

Hence, the points are $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$

Let the centre of circle be $(h, k)$ and radius be $r$.

$ \begin{aligned}Clearly,\, P M & =k \\ P N & =h \\ A M & =4, C N=3 \\ r^2 & =4^2+k^2 \\ r^2 & =3^2+h^2 \end{aligned} $

$ \begin{aligned} \because & 16+k^2=9+h^2 \\ & h^2-k^2-7=0 \end{aligned} $

Taking locus of $P(h, k)$, we get

$ x^2-y^2-7=0 $

Equation of tangent to the circle,

$ x^2+y^2=9 \text { is } y=m x+3 \sqrt{1+m^2} $

Tangent passes through the point $(3,4)$,

$ \begin{aligned} & (4-3 m)^2=9\left(1+m^2\right) \\ & \Rightarrow \quad 16+9 m^2-24 m=9+9 m^2 \\ & \Rightarrow \quad 24 m=7 \\ & \quad m=\frac{7}{24} \end{aligned} $

$ \text { } \begin{aligned} S & =x^2+y^2+2 k x+4 y-3=0 \\ C_1 & =(-k,-2) \text { radius } r_1=\sqrt{k^2+4+3} \\ S^{\prime} & =x^2+y^2-4 x+2 k y+9=0 \\ C_2 & =(2,-k) \text { radius }=\sqrt{4+k^2-9} \\ & =\sqrt{k^2-5} \end{aligned} $

Now, angle between two circles $S=0$ and $S^{\prime}=0$ is

$ \cos \left(180^{\circ}-\theta\right)=\frac{\left(r_1^2+r_2^2-\left(d^2\right)\right.}{2 r_1 r_2} $

where $r_1, r_2$ are the radii and $d$ is the distance between their centres.

$ \begin{aligned} & =\frac{k^2+7+k^2-5-(k+2)^2-\left(k-2)^2\right.}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & \Rightarrow-\cos \theta=\frac{2 k^2+2-\left(2 k^2+8\right)}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & =\frac{-6}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & \Rightarrow \quad \cos \theta=\frac{3}{\sqrt{k^2+7} \sqrt{k^2-5}}=\frac{3}{8} \\ & \Rightarrow \quad\left(k^2+7\right)\left(k^2-9\right)=64 \\ & \Rightarrow \quad k^4+2 k^2-99=0 \\ & \Rightarrow \quad\left(k^2+11\right)\left(k^2-9\right)=0 \Rightarrow k^2=9 \\ & \Rightarrow \quad k= \pm 3 \end{aligned} $

Centre of $S^{\prime}=0$ lies in I quadrant

$ [\therefore k=-3] $

Now, radical axis of $S=0$ and $S^{\prime}=0$ is

$ \Rightarrow \quad \begin{aligned} & x(2 k+4)+y(4-2 k)-12=0 \\ & -2 x+10 y-12=0 \\ & x-5 y+6=0 \end{aligned} $

The two given normal equations pass through the center of the circle. By solving the equations $2x - 3y + 4 = 0$ and $x + y - 3 = 0$, we find:

$ x = 1, \quad y = 2 $

Thus, the coordinates of the center of the circle are $(1, 2)$. Given that the point $(4, 6)$ lies on the circumference of the circle, we can calculate the radius as follows:

$ \text{Radius} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5 $

The circumference of the circle is then given by:

$ 2\pi r = 10\pi $

Coordinate of center of circle $x^2+y^2-2 x-4 y=0$ is $(1,2)$

$O M \perp A B$ and $M$, also $M$ is mid-point of $A B$.

Radius of circle $=\sqrt{(4-1)^2+(2-2)^2}=3$

Equation of line $P B$ is

$ \begin{aligned} & y-3=\left(\frac{2-3}{4-5}\right)(x-5) \\ & x-y=2 \\ & O M=\frac{|1-2-2|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \begin{aligned} A M & =\sqrt{O A^2-O M^2} \\ & =\sqrt{9-\frac{9}{2}}=\frac{3}{\sqrt{2}} \end{aligned} \end{aligned} $

$ \begin{aligned} & \qquad \begin{aligned} P B & =P A+2 A M \\ & =\sqrt{(5-4)^2+(3-2)^2}+2 \times \frac{3}{\sqrt{2}} \\ & =\sqrt{2}+\frac{6}{\sqrt{2}}=\frac{8}{\sqrt{2}} \end{aligned} \\ & \text { Now, } P A \times P B=\sqrt{2} \times \frac{8}{\sqrt{2}}=8 . \end{aligned} $

$P$ and $A(3,2)$ are conjugates points of the circle $x^2+y^2=4$.

Polar of $A$ is $3 x+2 y=4$

Polar of $A$ panes through $P$

$ 3 \alpha+2 \beta=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, point $P$ is on the line $2 x+y=1$

$ \Rightarrow \quad 2 \alpha+\beta=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ \alpha=-2 \text { and } \beta=5 $

Hence, $\alpha+\beta=-2+5=3$.

Center of the given circle is $(2,4)$ and radius of the given circle

$ \begin{aligned} & =\sqrt{4+16-16} \\ & =2 \end{aligned} $

As the perpendicular line from center bisects the chord,

$ A M=M B $

Slope of line $O M=\frac{4-3}{2-1}=1$

Slope of chord $A B=-1[\because A M \perp O M]$

Equation of chord $A B$ is

$ \begin{aligned} & y-3=-1(x-1) \\ & x+y=4 \\ & \frac{x}{4}+\frac{y}{4}=1 \end{aligned} $

$\therefore x$ intercept $=y$ intercept $=4$

$\therefore$ Required area of triangle

$=\frac{1}{2} \times 4 \times 4=8$ sq unit

If the two circle meets orthogonal, then

$ \begin{aligned} 2 g_1 g_2+2 f_1 f_2 & =c_1+c_2 \\ 2 a(-1)+a & =-8-14 \\ a & =22 \end{aligned} $

Center of two given circle are $(-22,-1)$ and $(1,-11)$

$\therefore$ Required distance

$ \begin{aligned} & =\sqrt{(1+22)^2+(-11+1)^2} \\ & =\sqrt{529+100} \\ & =\sqrt{629} \end{aligned} $

The given equation is:

$ x^2 + y^2 + 2x + 2y - 5 = 0 $

We need to consider the transformation when the axes are rotated by an angle $\alpha$. The equations for the transformed coordinates are:

$ x = X \cos \alpha - Y \sin \alpha $

$ y = X \sin \alpha + Y \cos \alpha $

Substituting these into the given equation, we get:

$ (X \cos \alpha - Y \sin \alpha)^2 + (X \sin \alpha + Y \cos \alpha)^2 + 2(X \cos \alpha - Y \sin \alpha) + 2(X \sin \alpha + Y \cos \alpha) - 5 = 0 $

Expanding and simplifying, the equation becomes:

$ X^2 + Y^2 + 2X \cos \alpha - 2Y \sin \alpha + 2X \sin \alpha + 2Y \cos \alpha - 5 = 0 $

Further simplifying, we have:

$ X^2 + Y^2 - 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0 $

For the transformed equation to have no linear terms, the coefficients of $X$ and $Y$ must be zero. Therefore, we set:

$\cos \alpha + \sin \alpha = 0$

$\cos \alpha - \sin \alpha = 0$

However, both equations cannot be simultaneously satisfied. Since these conditions are not possible together, there is no value of $\alpha$ that can make the transformed equation free of linear terms.

$ x^2+y^2=25 $

$ \begin{aligned} &\begin{aligned} R Q & =\sqrt{(3+4)^2+(4-3)^2} \\ & =\sqrt{49+1}=5 \sqrt{2} \\ O Q & =O R=5 \\ \cos \angle R O Q & =\frac{O R^2+O Q^2-R Q^2}{2 O R \cdot O Q} \\ & =\frac{25+25-50}{2 \times 5 \times 5}=0 \\ \Rightarrow \angle R O Q & =90^{\circ} \end{aligned}\\ &\text { Now, }\\ &\angle R P Q=\frac{1}{2} \angle R O Q=\frac{1}{2} \times 90^{\circ}=45^{\circ}=\frac{\pi}{4} \end{aligned} $

To determine the locus of the point of intersection of perpendicular tangents to the circle given by $x^2 + y^2 = 10$, we need to consider the concept of a director circle.

The director circle is a concentric circle whose radius is $\sqrt{2}$ times the radius of the original circle.

For the circle $x^2 + y^2 = 10$, the radius is $\sqrt{10}$. Thus, the radius of the director circle is:

$ \sqrt{2} \times \sqrt{10} = \sqrt{20} $

Thus, the equation of the director circle is:

$ x^2 + y^2 = 20 $

To find the equation of the normal to the circle $x^2 + y^2 - 4x + 6y - 4 = 0$ at the point $(1, 1)$, we first determine the circle's center.

The given equation can be rewritten as:

$ (x^2 - 4x) + (y^2 + 6y) = 4 $

Completing the square for both $x$ and $y$:

$ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 4 $

$ (x - 2)^2 + (y + 3)^2 = 17 $

Thus, the center of the circle is $(2, -3)$.

The normal line from $(1, 1)$ passes through the center $(2, -3)$. The slope of this normal line is calculated as follows:

$ \text{slope} = \frac{-3 - 1}{2 - 1} = -4 $

Using the point-slope form of a line with point $(1, 1)$:

$ y - 1 = -4(x - 1) $

Simplifying this equation:

$ y - 1 = -4x + 4 $

$ y = -4x + 5 $

In standard form, this can be written as:

$ 4x + y - 5 = 0 $

Thus, the equation of the normal is $4x + y - 5 = 0$.

Given equation of circle is

$ \begin{aligned} & \quad 2 x^2+2 y^2=9 \\ & \Rightarrow \quad 2\left(x^2+y^2\right)=9 \Rightarrow x^2+y^2=9 / 2 \\ & \Rightarrow \quad x^2+y^2=\left(\frac{3}{\sqrt{2}}\right)^2 \end{aligned} $

Now, the parametric equation of circle is

$ x=r \cos \theta, y=r \sin \theta $

$ \begin{aligned} & \therefore \quad x=\frac{3}{\sqrt{2}} \cos \theta \quad\left[\begin{array}{c} \because r^2=x^2+y^2 \\ r=\frac{\sqrt{3}}{2} \end{array}\right] \\ & \text { and } \quad y=\frac{3}{\sqrt{2}} \sin \theta \end{aligned} $

To find the angle between the circles $c_1$ and $c_2$, where:

$c_1: x^2 + y^2 - 4x - 6y - 3 = 0$

$c_2: x^2 + y^2 + 8x - 4y + 11 = 0$,

we can use the formula:

$ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} $

where $r_1$ and $r_2$ are the radii of the circles, and $d$ is the distance between their centers.

First, calculate the radii:

$ r_1 = \sqrt{2^2 + 3^2 + 3} = \sqrt{16} = 4 $

$ r_2 = \sqrt{4^2 + 2^2 - 11} = \sqrt{9} = 3 $

Next, determine the centers of the circles:

The center of $c_1$ is $(2, 3)$.

The center of $c_2$ is $(-4, 2)$.

Now, calculate the distance $d$ between the centers:

$ d = \sqrt{(-4 - 2)^2 + (2 - 3)^2} = \sqrt{36 + 1} = \sqrt{37} $

Substitute all values into the formula for the angle:

$ \cos \theta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \right| = \left| \frac{16 + 9 - 37}{2 \times 4 \times 3} \right| = \frac{1}{2} $

Thus, the angle $\theta$ is $\frac{\pi}{3}$.

$x^2+y^2-2 x+2 y+1=0$

is a circle with centre $(1,-1)$ and radius 1 When, chord are drawn from of the point $(1,0)$ on the circle the locus will be $x^2+y^2=4$.

For a given point $\left(x_1, y_1\right)$ on circle $x^2+y^2=4$ the equation of chord joining the point of tangent any of the tangent from $\left(x_1, y_1\right)$ to the circle is $x x_1+y y_1=4$ given original point

$ \Rightarrow x \cdot 1+y \cdot 0=4 \Rightarrow x=4 $

Thus, locus of poles of these chords is a vertical line.

$x^2+y^2-14 x+6 y+33=0\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ x^2+y^2+30 x-2 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

using Eq. (i), we can write it as

$ \begin{array}{r} (x-7)^2-49+(y+3)^2-9+33=0 \\ (x-7)^2+(y+3)^2-25=0 \end{array} $

So, centre $(7,-3)$ and radius $=5$ using Eq. (ii) we can write it as

$ \begin{aligned} (x+15)^2-225+(y-1)^2-1+1 & =0 \\ (x+15)^2+(y-1)^2 & =225 \end{aligned} $

So, centre $(-15,1)$ and radius $=15$

Now, A(External centre similitude)

$ =\frac{R_2 C_1-R_1 C_2}{R_2-R_1} $

And B (Internal centre similitude) $=\frac{R_2 C_1+R_1 C_2}{R_2+R_1}$, where $C_1$ and $C_2$ are

centers and $R_1$ and $R_2$ are their radii.

$ \begin{aligned} A & =\frac{15(7,-3)-5(-15,1)}{15-5} \\ & =\frac{(105,-45)+(75,-5)}{10} \\ & =\frac{(180,-50)}{10}=(18,-5) \\ B & =\frac{15(7,-3)+(5)(-15,1)}{15+5} \\ & =\frac{(105,-45)+(-75,5)}{20} \\ & =\frac{(30,-40)}{20}=(1.5,-2) \end{aligned} $

Finally mid-point $M$ of line segment

$ \begin{aligned} M & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\ & =\left(\frac{18+1.5}{2}, \frac{-5+(-2)}{2}\right) \\ M & =\left(\frac{19.5}{2}, \frac{-7}{2}\right)=\left(\frac{39}{4}, \frac{-7}{2}\right) \end{aligned} $

Given that $C_1$ is a circle centered at $O(0,0)$ with a radius of 4, and $C_2$ is a variable circle with a center at $(\alpha, \beta)$ and a radius of 5.

The equations of the circles are:

$C_1: x^2 + y^2 = 16$

$C_2: (x - \alpha)^2 + (y - \beta)^2 = 25$

For the common chord to exist and have maximum length, the distance between the centers ($\sqrt{\alpha^2 + \beta^2}$) is:

$ \sqrt{5^2 - 4^2} = 3 $

The slope of the common chord is given as $\frac{3}{4}$. Therefore, the equation of the chord can be written as:

$ 3x - 4y = 0 \quad ...(i) $

Since the distance between the points $(0,0)$ and $(\alpha, \beta)$ should be 3, we find:

$ \left| \frac{3\alpha - 4\beta}{\sqrt{3^2 + 4^2}} \right| = 3 $

Simplifying further, we have:

$ 3\alpha - 4\beta \pm 15 = 0 \quad ...(ii) $

Given that the product of the slopes of perpendicular lines is -1, and the line connecting the centers $O$ and $(\alpha, \beta)$ is perpendicular to the chord, we have:

$ \frac{3}{4} \times \frac{\beta}{\alpha} = -1 $

Which simplifies to:

$ 3\beta + 4\alpha = 0 \quad ...(iii) $

By solving equations (ii) and (iii), we find:

$\alpha = -\frac{9}{5}$

$\beta = \frac{12}{5}$

Therefore, $\alpha + \beta = \frac{3}{5}$.

To determine the radius $a$ for which the tangents from the point $(10, 4)$ to the circle $x^2 + y^2 = a^2$ are perpendicular, we use the property that tangents from a given point $(x_1, y_1)$ to a circle are perpendicular if:

$ x_1^2 + y_1^2 = 2r^2 $

Given that the point is $(10, 4)$ and the circle is defined by $x^2 + y^2 = a^2$, we set $x_1 = 10$, $y_1 = 4$, and $r = a$. Plugging these into the formula gives:

$ 10^2 + 4^2 = 2a^2 $

Calculating further:

$ 100 + 16 = 2a^2 $

$ 116 = 2a^2 $

Dividing by 2:

$ a^2 = \frac{116}{2} = 58 $

Thus, the radius $a$ is:

$ a = \sqrt{58} $

Given the circle $ x^2 + y^2 = 36 $, we have a circle $ S_2 $ centered at $(0, 0)$ with a radius of 6. The radical axis of two orthogonal circles is given by $ x - 4 = 0 $.

For two circles to be orthogonal, the condition is:

$ 2(g_1g_2 + f_1f_2) = c_1 + c_2 $

where $(g_1, f_1)$ and $(g_2, f_2)$ are the centers of the circles, and $c_1$ and $c_2$ are the constants in their respective equation forms $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$.

Let's denote the other circle as $ S_1 $. According to the given problem, the radical axis is:

$ x - 4 = 0 $

The equation of the other circle $ S_1 $ can be expressed as:

$ S_1 - S_2 = k(x - 4) $

So:

$ S_1 = S_2 + k(x - 4) $

This translates to:

$ S_1 = x^2 + y^2 - 36 + kx - 4k = 0 $

Substituting into the orthogonality condition:

$ c_1 + c_2 = 0 $

Hence:

$ -4k - 36 - 36 = 0 $

Solving for $ k $:

$ -4k - 72 = 0 \Rightarrow k = -18 $

Therefore, the equation of the other circle $ S_1 $ becomes:

$ x^2 + y^2 - 18x + 36 = 0 $

The center of this circle is $(9, 0)$.

Let $ Q = (x_1, y_1) $ and $ A = (4, 4) $.

The circle is defined by the equation $ x^2 + y^2 = 9 $.

The coordinates of the point $ P $, which divides the line segment $ QA $ in the ratio $ m:n = 1:2 $, are determined by the formula:

$ P\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right) $

Substituting the given ratio and coordinates:

$ P\left(\frac{1 \cdot 4 + 2 \cdot x_1}{1+2}, \frac{1 \cdot 4 + 2 \cdot y_1}{1+2}\right) \Rightarrow P\left(\frac{4 + 2x_1}{3}, \frac{4 + 2y_1}{3}\right) \tag{i} $

Since $ Q $ is on the circle $ x^2 + y^2 = 9 $, we have:

$ x_1^2 + y_1^2 = 9 $

Using parametric equations for the circle, let $ x_1 = 3 \cos \theta $ and $ y_1 = 3 \sin \theta $. Substituting into the expression for $ P $ from Eq. (i):

$ P\left(\frac{4 + 6 \cos \theta}{3}, \frac{4 + 6 \sin \theta}{3}\right) \Rightarrow P\left(\frac{4}{3} + 2 \cos \theta, \frac{4}{3} + 2 \sin \theta\right) $

Let:

$ x = \frac{4}{3} + 2 \cos \theta \quad \Rightarrow \quad \cos \theta = \frac{x - \frac{4}{3}}{2} $

$ y = \frac{4}{3} + 2 \sin \theta \quad \Rightarrow \quad \sin \theta = \frac{y - \frac{4}{3}}{2} $

Substitute into the trigonometric identity $ \sin^2 \theta + \cos^2 \theta = 1 $:

$ \left(\frac{x - \frac{4}{3}}{2}\right)^2 + \left(\frac{y - \frac{4}{3}}{2}\right)^2 = 1 $

Simplifying further:

$ \left(x - \frac{4}{3}\right)^2 + \left(y - \frac{4}{3}\right)^2 = 4 $

This equation represents a circle with center $\left(\frac{4}{3}, \frac{4}{3}\right)$ and radius 2. Thus, the perimeter of this circle is:

$ 2 \pi \times \text{radius} = 4 \pi $

Radius of circle is 3 units and which touches internally the circle $\Rightarrow C \equiv x^2+y^2-4 x-6 x-12=0$ at points $(-1,1)$

Centre of circle $C$ is $(2,3)$ and radius is 5 units

If $C_2(h, K)$ is the centre of circle of radius 3. Which touches circle $C$ internally at $(1,-1)$, then

$ \Rightarrow C_2=3 \text { and } \Rightarrow C_1, C_2=R-r=5-3=2 $

Thus, $C_2(h, K)$ divides $C$ in the ratio $2: 3$ internally therefore,

$ \begin{aligned} & \Rightarrow h=\frac{2(-1)+3 \times 2}{2+3}=\frac{4}{5} \\ & \Rightarrow k=\frac{2(-1)+3 \times 3}{2+3}=\frac{7}{5} \end{aligned} $

Hence, equations of circle $C_2$ is

$ \begin{aligned} & \Rightarrow\left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\ & \Rightarrow x^2+\frac{16}{25}-\frac{8 x}{5}+y^2+\frac{49}{25}-\frac{14 y}{5}=9 \\ & \Rightarrow x^2+y^2-\frac{8 x}{5}-\frac{14 y}{5}-\frac{32}{5}=0 \end{aligned} $

On comparing with given equation of circle

$ x^2+y^2+p x+q y+r=0 $

We get, $p=\frac{-8}{5}, q=\frac{-14}{5}, r=\frac{-32}{5}$

$ \begin{aligned} & \Rightarrow p+q-r=\frac{-8}{5}-\frac{14}{5}+\frac{32}{5}=2 \\ & \Rightarrow p+q-r=2 \end{aligned} $

Equation of circle

$ x^2+y^2-6 x-6 y+17=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and equation of line

$ x^2-3 x y-3 x+9 y=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

$ \begin{aligned} &\text { From Eq. (ii), we get }\\ &\begin{aligned} \Rightarrow & & x^2-3 x y-3 x+9 y & =0 \\ \Rightarrow & & x^2-3 x-3 x y+9 y & =0 \\ \Rightarrow & & x(x-3)-3 y(x-3) & =0 \\ \Rightarrow & & (x-3)(x-3 y) & =0 \\ \Rightarrow & & x=3, x & =3 y \\ \Rightarrow & & x=3, y & =1 \\ & & R & =\sqrt{9+9-17} \\ & & R & =\sqrt{1}=1 \end{aligned} \end{aligned} $

According to the given condition,

$ \begin{array}{cc} \Rightarrow & C_1 C_2=R+r \\ \Rightarrow & \sqrt{(3-3)^2+(1+3)^2}=1+r \\ \Rightarrow & 4=1+r \\ \Rightarrow & r=3 \end{array} $

Equation of circle having coordinates $(3,1)$ and radius 3 is

$ \begin{aligned} & \Rightarrow(x-3)^2+(y-1)^2=9 \\ & \Rightarrow x^2+9-6 x+y^2+1-2 y=9 \\ & \Rightarrow x^2+y^2-6 x-2 y+1=0 \end{aligned} $

Above equation is equation of circle which touches the circle.

We have, equation of straight line

$ 9 x+y-28=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and equation of circle

$ C \equiv 2 x^2+2 y^2-3 x+5 y-7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Let ( $x_1, y_1$ ) be the required point, line (i) must coincide with the polar of $\left(x_1, y_1\right)$ whose equation is

$ \begin{aligned} & \Rightarrow 2 x x_1+2 y y_1-\frac{3}{2}\left(x+x_1\right)+\frac{5}{2}\left(y+y_1\right)-7 \\ & =0 \\ & \begin{aligned} \Rightarrow & x\left(4 x_1-3\right)+y\left(4 y_1+5\right)-3 x_1 \end{aligned}+5 y_1-14=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Since, Eqs. (i) and (iii) are same, we have

$ \frac{4 x_1-3}{9}=\frac{y_1+5}{1}=\frac{-3 x_1+5 y_1-14}{-28} $

$ \begin{aligned} & \Rightarrow \quad x_1=9 y_1+12 \\ & \Rightarrow 3 x_1-117 y_1=126 \end{aligned} $

On solving, we get

$ x=3 \Rightarrow y=-1 $

So, the required point is $(3,-1)$.

We have, equations of lines

$ \begin{array}{ll} \Rightarrow & L_1 \equiv x+y=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & L_2 \equiv x-y=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Let the cordinate of centre of circle be $(g, f)$ which touches the line i.e. it will satisfy the equations of lines $L_1$ and $L_2$.

$ \begin{array}{ll} \Rightarrow & g+f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \Rightarrow & g-f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $

On solving Eqs. (iii) and (iv), we get (g. $f$ ) $=(2,0)$

The center of the required circle will lie on bisector of acute angle between these tangents i.e. on.

$ \frac{x+y-2}{\sqrt{2}}=\frac{x-y-2}{\sqrt{2}} \text { i.e. } $

Let it be ( $h, 0$ ) where $h$ is (positive). If $r$ be the radius, then it touches the circle $x^2+y^2=1$ externally

According to the condition,

$ \begin{aligned} & C_1 C_2 =r_1+r_2\\& h =1+r \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(v)\\ \therefore \quad h & =2 \pm \sqrt{2 r}=1+r \end{aligned} $

On solving, we get, $r=\sqrt{2}-1$

Hence, the equation of circle with center $(\sqrt{2}, 0)$ and radius $(\sqrt{2}-1)$ is

$ \begin{aligned} & \Rightarrow(x-\sqrt{2})^2+y^2=(\sqrt{2}-1)^2 \\ & \Rightarrow(x-\sqrt{2})^2+y^2=3-2 \sqrt{2} \end{aligned} $

We have two given circles:

$ \begin{aligned} & S_1 \equiv x^2 + y^2 + 2gx + 2fy + c = 0 \\ & S_2 \equiv 2x^2 + 2y^2 + 3x + 8y + 2c = 0 \end{aligned} $

The equation of the second circle can be simplified by dividing by 2:

$ S_2 \equiv x^2 + y^2 + \frac{3}{2}x + 4y + c = 0 $

The radical axis of the circles $ S_1 = 0 $ and $ S_2 = 0 $ is determined by subtracting one equation from the other:

$ \begin{aligned} & S_1 - S_2 = 0 \\ & \Rightarrow \left(x^2 + y^2 + 2gx + 2fy + c\right) - \left(x^2 + y^2 + \frac{3}{2}x + 4y + c\right) = 0 \\ & \Rightarrow 2gx + 2fy - \frac{3}{2}x - 4y = 0 \\ & \Rightarrow x(g - \frac{3}{4}) + y(f - 2) = 0 \end{aligned} $

This radical axis touches the circle:

$ S_3 \equiv x^2 + y^2 + 2x + 2y + 1 = 0 $

The center of this circle is $(-1, -1)$ and it has a radius of 1. For tangency, the distance from the center to the radical axis must equal the radius, i.e., $p = r$:

$ \frac{-1(g - \frac{3}{4}) - (f - 2)}{\sqrt{(g - \frac{3}{4})^2 + (f - 2)^2}} = 1 $

Squaring both sides, we arrive at:

$ \begin{aligned} & (g - \frac{3}{4})^2 + (f - 2)^2 + 2(g - \frac{3}{4})(f - 2) \\ & = (g - \frac{3}{4})^2 + (f - 2)^2 \\ & \therefore 2(g - \frac{3}{4})(f - 2) = 0 \end{aligned} $

This implies either $g - \frac{3}{4} = 0$ or $f - 2 = 0$:

$ g = \frac{3}{4} \quad \text{or} \quad f = 2 $

The point of intersection of the given linear equation is $(4,3)$.

So, center of the circle will be $(4,3)$.

$ \begin{aligned} & \therefore S \equiv x^2+y^2-8 x-6 y-11=0 \\ & \begin{aligned} \text { Radius } & =\sqrt{(-4)^2+(-3)^2-(-11)} \\ & =\sqrt{36}=6 \end{aligned} \end{aligned} $

The most rightward point of circle will be $(-2,3)$. The $x$-coordinate of this point is same as the $x$-coordinate of the point $P(-2,-2)$.

So, the line $x=-2$ will be a tangent from $P$ and it will touch the circle at $(-2 ; 3)$.

Clearly, $\triangle C P R \cong \triangle C P Q$ and $\triangle C P R$ is a right-angled triangle.

$\therefore$ Area of quadrilateral $P Q C R=2 \times$ Area of $\triangle C P R$

$ \begin{aligned} & =2 \times\left(\frac{1}{2} \times 5 \times 6\right) \\ & =30 \end{aligned} $

Given, if the inverse point of the point $(-1,+1)$ w.r.t. the

Circle $x^2+y^2-2 x+2 y-1=0$ is $(p, q)$ then, $p^2+q^2=$ ?

Now, $x^2-2 x+y^2+2 y=1$

1 Add or subtract

$ \begin{aligned} & \left(x^2-2 x+1-1\right)+\left(y^2+2 y+1-1\right)=1 \\ & \left(x^2-2 x+1\right)-1+\left(y^2+2 y+1\right)-1=1 \\ & (x-1)^2-1+(y+1)^2-1=1 \\ & (x-1)^2+(y+1)^2=3 \end{aligned} $

So, center of circle $(1,-1)$ and Radius $=\sqrt{3}$

Now, find the distance from the point $(-1,1)$ to the center of the circle.

$ \begin{aligned} & \sqrt{(-1-1)^2+(1-(-1))^2} \\ &=\sqrt{(-2)^2+(2)^2}=2 \sqrt{2} \end{aligned} $

Now, determine the inverse point $(p, q)$ of $(-1,1)$ with respect to the circle lies on the same line passing through the point.

Here, $\left(x_1, y_1\right)=(-1,1)$

$ \begin{aligned} (a, b) & =(1,-1), \\ r^2 & =3 \end{aligned} $

Formula for inverse point

$ \begin{aligned} & p=\left(a+\left\{r^2+\left(x_1-a\right) /\left(\left(x_1-a\right)^2+\left(y_1-b\right)^2\right)\right\}\right) \\ & q=\left(b+\left\{r^2\left(y_1-b\right) /\left(\left(x_1-a\right)^2+\left(y_1-b\right)^2\right)\right\}\right) \\ & \begin{aligned} p=1 & +\frac{3 x-2}{8}=\frac{8-6}{8}=\frac{2}{8}=\frac{1}{4} \\ q & =-1+\frac{3(1+1)}{4+4} \\ & =-1+\frac{6}{8}=-\frac{1}{4} \end{aligned} \\ & p^2+q^2=\left(\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)^2=\frac{1}{8} \end{aligned} $

Mid-point $(a, b)$ of the chord. $(2 x-y+3=0)$ of the circle. $x^2+y^2+6 x-4 y+4=0$

$ \begin{aligned} & \begin{array}{l} \left(x^2+6 x\right)+\left(y^2-4 y\right)+4=0 \\ \left(x^2+9-9+6 x\right)+\left(y^2+4-4-4 y\right)+4=0 \\ \left(x^2+9+9+6 x\right)-9+\left(y^2+4-4 y\right)=0 \\ \qquad(x+3)^2+(y-2)^2=9 \end{array} \\ & \text { Center }(-3,2) \text { and Radius }=3 \end{aligned} $

The chord equation, $2 x-y+3=0$

The mid-point $(a, b)$ is perpendicular bisector of the chord which passes thought center of the circle $(-3,2)$. The equation of the perpendicular bisectord the chord can be found using the given chord equation and the center of the circle rewriting chord equation.

$ y=2 x+3 $

Slope of this line is 2 . Therefore, the slope of perpendicular bisector is $-\frac{1}{2}$

Now, center $(-3,2) m=\frac{-1}{2}$

$ \begin{aligned} y-2 & =-\frac{1}{2}(x+3) \\ y & =-\frac{1}{2} x+\frac{1}{2} \end{aligned} $

The mid-point $(a, b)$ of the chord is the intersection. Point of the original chord $2 x-y \neq 3=0$ and perpendicular bisettr $y=\frac{-1}{2} x+\frac{1}{2}$.

On substitute the value,

$ 2 x-\left(-\frac{1}{2} x+\frac{1}{2}\right)+3=0 $

On solving $x=11$ and $y=1$

So, the mid-point $(a, b)$ is $(-1,1)$.

Calculate $2 a+3 b$

$ \Rightarrow 2(-1)+3(1) \Rightarrow 1 $

Therefore, $2 a+3 b=1$.

Given, first circle,

$ x^2+y^2-6 x+4 y+9=0 $

By complete the square,

$ (x-3)^2+(y+2)^2=4 $

Center $C_1(3,-2)$, Radius $\left(R_1\right)=2$

Similarly, for second circle.

$ x^2+y^2+2 x-2 y+1=0 $

By complete the square,

$ (x+1)^2+(y-1)^2=1 $

Center $C_2(-1,1)$, Radius $R_2=1$

$ \begin{aligned} & \text { Distance between } C_1 \text { and } C_2 \\ & \begin{aligned} d=\sqrt{(3-(-1))^2+(-2-1)^2}=\sqrt{4^2+(-3)^2} \,\,=5\\ \end{aligned} \\ & \begin{aligned} \text { Length of tangent } & =\sqrt{d^2-\left(R_1-R_2\right)^2} \\ & =\sqrt{5^2-(2-1)^2} \\ & =\sqrt{25-1} \\ & =\sqrt{24} \\ & =2 \sqrt{6} \end{aligned} \end{aligned} $

Therefore, the length of the common tangent is $2 \sqrt{6}$ unit.

Given, 1st circle

$ =x^2+y^2-4 x-4 y+7=0 $

2nd circle $=x^2+y^2+4 x-4 y+6=0$

3rd circle $=x^2+y^2+4 x+4 y+5=0$

Now, find the center and radii of the given circles.

1st circle $x^2+y^2-4 x-4 y+7=0$

By complete the square

$ \begin{aligned} & (x-2)^2+(y-2)^2=2^2+2^2-7 \\ & (x-2)^2+(y-2)^2=1 \end{aligned} $

Center $C_1(2,2)$ Radius $R_1=1$

Similarly, 2nd Circle $(x+2)^2+(y-2)^2=2$

Center $C_2(-2,2)$ and Radius $R_2=\sqrt{2}$

3rd circle $(x+2)^2+(y+2)^2=3$

$ C_3(-2,-2), R_3=\sqrt{3} $

Now, let the Radius $R$ and center $(a, b)$ be the circle which cut each of given circles orthogonally, then the below condition must hold.

$ \begin{aligned} & (a-2)^2+(b-2)^2=R^2+1 \\ & (a+2)^2+(b-2)^2=R^2+2 \\ & (a+2)^2+(b+2)^2=R^2+3 \end{aligned} $

On subtract Eq. (i) from Eq. (ii), we get

$ a=\frac{1}{8} $

On subtracting Eq. (i) from Eq. (iii), we get

$ b=\frac{1}{8} $

On substitute $a$ and $b$ in Eq. (i)

$ \begin{aligned} &\left(\frac{1}{8}-2\right)^2+\left(\frac{1}{8}-2\right)^2=R^2+1 \\ &\left(\frac{-15}{8}\right)^2+\left(\frac{-15}{8}\right)^2=R^2+1 \\ & \frac{450}{64}=R^2+1 \\ & R^2=\frac{193}{32} \\ & R=\sqrt{\frac{193}{32}} \\ & R=\frac{\sqrt{193}}{4 \sqrt{2}} \end{aligned} $

Therefore, the radius of the circle that cuts all the given circle orthogonally is $\frac{\sqrt{193}}{4 \sqrt{2}}$ unit.

In $\triangle A P B$, we have

$\begin{aligned} & m_1=\text { Slope of } A P=\frac{y-3}{x-2} \\ & m_2=\text { Slope of } B P=\frac{y-1}{x+1}\end{aligned}$

Now, $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\begin{aligned} & \tan 90^{\circ}=\left|\frac{\frac{(y-3)}{(x-2)}-\frac{(y-1)}{(x+1)}}{1+\frac{(y-3)}{(x-2)} \frac{(y-1)}{(x+1)}}\right| \\ & \quad=\left|\frac{(y-3)(x+1)-(y-1)(x-2)}{(x-2)(x+1)+(y-3)(y-1)}\right| \\ & \quad \frac{1}{\infty}=0=\frac{(x-2)(x+1)+(y-3)(y-1)}{(y-3)(x+1)-(y-1)(x-2)} \\ & \Rightarrow x^2-x-2+y^2-4 y+3=0 \\ & \Rightarrow x^2-x+y^2-4 y+1=0\end{aligned}$

Given the equation of the circle:

$ x^2 + y^2 - 6x - 8y - 11 = 0 $

we can determine the center of the circle, $ C $, which is $ (3, 4) $, and the radius, $ r $, as follows:

$ r = \sqrt{9 + 16 + 11} = \sqrt{36} = 6 $

Now, let's consider the point $ P(15, 9) $.

The distance between the point $ P $ and the center $ C $ is calculated as:

$ CP = \sqrt{(15 - 3)^2 + (9 - 4)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $

To find the largest possible distance from the point $ P $ to any point on the circle, we add the radius to $ CP $:

$ \text{Greatest distance from } P \text{ to the circle} = CP + r = 13 + 6 = 19 $

Given the circle equation:

$ x^2 + y^2 - 8x - 12y + \alpha = 0 $

And the point $ P(6, 6) $ is an interior point of the circle. To determine the range of $\alpha$, we perform the following steps:

Point $ P(6, 6) $ lies inside the circle, thus:

$ 6^2 + 6^2 - 8 \cdot 6 - 12 \cdot 6 + \alpha < 0 $

$ 36 + 36 - 48 - 72 + \alpha < 0 $

Simplifying:

$ 72 - 48 - 72 + \alpha < 0 $

$ \alpha < 48 $

To ensure the circle does not touch the coordinate axes and remains in the first quadrant, the radius must be less than both the x-coordinate and y-coordinate of the center. The center of the circle $(h, k)$ is found by completing the square:

$ h = \frac{8}{2} = 4, \quad k = \frac{12}{2} = 6 $

Thus, the center is at $(4, 6)$ and the radius $ \sqrt{4^2 + 6^2 - \alpha} < 4 $ and also $ < 6 $:

$ \sqrt{52 - \alpha} < 4 \quad \text{and} \quad \sqrt{52 - \alpha} < 6 $

Solving these gives:

$ 52 - \alpha < 16 \quad \text{and} \quad 52 - \alpha < 36 $

Which simplifies to:

$ \alpha > 36 $

Combining the inequalities from steps 1 and 2, we have:

$ 36 < \alpha < 48 $

To find the equation of the circle whose diameter is the common chord of the circles given by:

$ x^2 + y^2 - 6x - 7 = 0 \quad (i) $

$ x^2 + y^2 - 10x + 16 = 0 \quad (ii) $

First, we find the common chord by subtracting equation (i) from equation (ii):

$ \begin{aligned} & \quad x^2 + y^2 - 10x + 16 - (x^2 + y^2 - 6x - 7) = 0, \\ & \Rightarrow -4x + 23 = 0, \\ & \Rightarrow x = \frac{23}{4}. \end{aligned} $

Substitute $ x = \frac{23}{4} $ back into either equation (let’s use equation (ii)) to find the $ y $-coordinates of the intersection points:

$ \begin{aligned} & x^2 + y^2 - 10x = -16, \\ & \left(\frac{23}{4}\right)^2 + y^2 - 10 \cdot \frac{23}{4} = -16, \\ & y = \frac{\pm 3 \sqrt{15}}{4}. \end{aligned} $

The points of intersection are $\left(\frac{23}{4}, \frac{-3 \sqrt{15}}{4}\right)$ and $\left(\frac{23}{4}, \frac{3 \sqrt{15}}{4}\right)$.

The midpoint of the chord (diameter) is:

$ \left(\frac{23}{4}, 0\right) $

The radius is the distance from this midpoint to either endpoint:

$ \begin{aligned} d &= \sqrt{\left(\frac{23}{4} - \frac{23}{4}\right)^2 + \left(0 - \frac{3 \sqrt{15}}{4}\right)^2} \\ &= \frac{3 \sqrt{15}}{4} \end{aligned} $

Thus, the equation of the circle is:

$ \begin{aligned} & \left(x - \frac{23}{4}\right)^2 + y^2 = \left(\frac{3 \sqrt{15}}{4}\right)^2, \\ & \Rightarrow \left(x - \frac{23}{4}\right)^2 + y^2 = \frac{135}{16}, \\ & \Rightarrow 16\left(x^2 - \frac{23}{2}x\right) + 16y^2 = 135, \\ & \Rightarrow 8x^2 + 8y^2 - 92x + 197 = 0. \end{aligned} $

This is the equation of the circle whose diameter is the common chord of the given circles.

Let mid-point be $P(h, k)$

and origin $O(0,0)$ is also the centre of the circle.

Since, chord making right angle at origin and $P$ is mid-point, $O P$ will bisect the right angle.

$O P=r \cos 45^{\circ}$, where $r$ is radius of given circle.

$ \begin{array}{ll} \Rightarrow & \sqrt{(h-0)^2+(k-0)^2}=5 \times \frac{1}{\sqrt{2}} \\ \Rightarrow & h^2+k^2=\left(\frac{5}{\sqrt{2}}\right)^2=\frac{25}{2} \\ \therefore & \quad O P=\frac{5}{\sqrt{2}} \end{array} $

Let the coordinates of $P$ be $\left(x_1, y_1\right)$

$ O P=\sqrt{x_1^2+y_1^2}=\frac{5}{\sqrt{2}} $

Let $S_1: x^2+y^2+2 x+3 y+1=0$

$ \begin{aligned} & S_2: x^2+y^2+x-y+3=0 \\ & S_3: x^2+y^2-3 x+2 y+5=0 \end{aligned} $

The radical axis of the circles will be

$ \begin{array}{l} & S_1-S_2=0 \\ \Rightarrow & x+4 y-2=0 \quad \ldots \text { (i) } \\ \text { and } & S_2-S_3=0 \\ & 4 x-3 y-2=0 \quad \ldots \text { (ii) } \\ \text { and } & S_3-S_1=0 \\ \Rightarrow & -5 x-y+4=0 \quad \ldots \text { (iii) } \end{array} $

On multiply Eqs. (i) by 4 and then subtract from (ii), we get

$ y=\frac{6}{19} $

On substitute this value in Eq. (i), we get

$ x=\frac{14}{19} $

Since, $x$ and $y$ values also satisfies Eqs. (iii).

Thus, the radical centre of three circles is $\left(\frac{14}{19}, \frac{6}{19}\right)$.

In, figure $B A$ and $B C$ are tangents $y$ the circle. So, $O B$ is the angle bisecton of $\angle B$.

$ \text { Then, } \begin{aligned} \angle B & =60^{\circ} \\ \angle O B D & =\frac{60}{2}=30^{\circ} \end{aligned}$

$ \text { In } \triangle O B D, \angle O D B=90^{\circ} $

$ \begin{aligned} \tan 30^{\circ} & =\frac{O D}{B D} \\ \frac{1}{3} & =\frac{r}{\frac{a}{2}}=\frac{2 r}{a} \\ r & =\frac{a}{2 \sqrt{3}} \end{aligned} $

Now, diagonal of square $=$ diameter of circle

$ =2 r=2 \times \frac{a}{2 \sqrt{3}}=\frac{a}{\sqrt{3}} $

Now, area of square $=\frac{1}{2}(2 r)^2$

$ =\frac{1}{2} \times\left(\frac{a}{\sqrt{3}}\right)^2=\frac{a^2}{6} $

Slope of $A P=\frac{k}{h-1}$

Slope of $B P=\frac{k-1}{h}$

$ \begin{aligned} & \tan 45^{\circ}=\left|\frac{\frac{k}{h-1}-\frac{k-1}{h}}{1+\frac{k(k-1)}{h(h-1)}}\right| \\ & \pm 1=\frac{k h-(k-1)(h-1)}{h(h-1)+k(k-1)} \\ & \pm 1=\frac{k h-k h+h+k-1}{h^2-h+k^2-k} \\ & \Rightarrow \pm\left(h^2-h+k^2-k\right)=h+k-1 \end{aligned} $

If, $h+k-1=-\left(h^2-h+k^2-k\right)$,

$\Rightarrow h+k-1=-h^2+h-k^2+k$ $h^2+k^2-1$

If $h+k-1=h^2-h+k^2-12$

$ \Rightarrow h^2+k^2-2 h-2 k+1=0 $

Equation of locus of $P$ is $\left(x^2+y^2-1\right)=0$

or $x^2+y^2-2 x-2 y+1=0$

Hence, equation of locus of $p$ is

$ \begin{aligned} & \left(x^2+y^2-1\right)\left(x^2+y^2-2 x-2 y+1\right)=0 \\ & x \neq 0,1 \end{aligned} $

Let centre of circle be $(h, k)$.

Centre of circle lies on $2 x+y+3=0$

$ 2 h+k+3=0 $

The two tangents $3 x+4 y-18=0$ and $3 x+4 y+2=0$ are parallel as their slopes are equal

$ 2 r=\frac{|2+18|}{\sqrt{3^2+4^2}}=\frac{20}{5}=4 $

$ \therefore \quad r=2 \text { units } $

Now, $r=\frac{|3 h+4 k-18|}{\sqrt{3^2+4^2}}=2$

$ \begin{array}{ll} \Rightarrow & 3 h+4 k-18= \pm 10 \\ & 3 h+4 k-18+10=0 \\ \text { or } & 3 h+4 k-18-10=0 \\ \Rightarrow & 3 h+4 k-8=0 \\ \text { Or } & 3 h+4 k-28=0 \end{array} $

From Eqs. (i) and (ii), we get $h=-4$, $k=5$

From Eqs. (i) and (iii), we get $h=-8$, $k=13$

Neglecting $h=-8$ and $k=13$

because distance from centre to tangent $\neq 2$

$\therefore$ Centre of circle is $(-4,5)$

$\therefore$ Equation of circle is

$ \begin{array}{r} (x+4)^2+(y-5)^2=2^2 \\ x^2+y^2+8 x-10 y+16+25=4 \\ x^2+y^2+8 x-10 y+37=0 \end{array} $

The power of the point $(4, 2)$ with respect to the given circle is 9.

To find the circle's equation, use the power of the point:

$ 16 + 4 - 8\alpha + 12 + \alpha^2 - 16 = 9 $

Solving this equation gives:

$ \alpha^2 - 8\alpha + 7 = 0 $

which results in:

$ \alpha = 1 \quad \text{or} \quad \alpha = 7 $

Thus, the possible equations of the circles are:

$x^2 + y^2 - 2x + 6y - 15 = 0$

$x^2 + y^2 - 14x + 6y + 33 = 0$

For the equation $x^2 + y^2 - 2x + 6y - 15 = 0$:

Length of intercepts on the X-axis:

$ 2\sqrt{g^2 - c} = 2\sqrt{1 + 15} = 2\sqrt{16} = 8 $

Length of intercepts on the Y-axis:

$ 2\sqrt{f^2 - c} = 2\sqrt{9 + 15} = 2\sqrt{24} = 4\sqrt{6} $

For the equation $x^2 + y^2 - 14x + 6y + 33 = 0$:

Length of intercept on the X-axis:

$ 2\sqrt{g^2 - c} = 2\sqrt{49 - 33} = 2\sqrt{16} = 8 $

Length of intercept on the Y-axis is not possible (as the term under the square root is negative).

Therefore, the sum of the lengths of all possible intercepts is:

$ 8 + 8 + 4\sqrt{6} = 16 + 4\sqrt{6} $

We have,

Equation of circle is

$ \begin{aligned} & x^2+y^2-4 x-6 y+9=0 \\ & \text { Centre of circle }=(2,3), r=2 \end{aligned} $

As the line intersects the circle at 2 points, perpendicular distance from centre to line in less than radius

$ \begin{aligned} & 0<\frac{|12+24+\alpha|}{\sqrt{6^2+8^2}}<2 \\ & 0<|36+\alpha|<20 \\ & \Rightarrow \text { Either } 0<36+\alpha<-16 \\ & \text { or } \quad 0<-(36+\alpha)<20 \end{aligned} $

$\Rightarrow$ Either $-36<\alpha<-16$, the possible values of $\alpha$ are $-32-24[\because \alpha$ is a multiple of 8]

If $-36>\alpha>-56$, then possible values of $\alpha$ are $-40,-48$.

Hence, the number of elements is set $s=4$.

$C_1$ : Centre of circle

$ \begin{aligned} & \quad x^2+y^2-8 x-8 y+28=0 \text { is }(4,4) \\ & r_1=\sqrt{16+16-28}=2 \end{aligned} $

${ }C_2$ : Centre of circle

$ \begin{aligned} & x^2+y^2-8 x-6 y+25-\alpha^2=0 \text { is }(4,3) \\ & r_2=\sqrt{16+9-25+\alpha^2}=\alpha \end{aligned} $

As the two circles have one common tangent, then the two circle coincides internally at point

$ \begin{aligned} C_1 C_2 & =\left|r_1-r_2\right| \sqrt{(4-4)^2+(4-3)^2}=|2-\alpha| \\ 1 & =|2-\alpha| \Rightarrow 2-\alpha= \pm 1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2-\alpha=1 \text { or } 2-\alpha=-1 \\ & \Rightarrow \quad \alpha=1 \text { or } \alpha=3 \end{aligned} $

The equation of the circle passing through the intersection of two given circles can be described by:

$ x^2 - 2x + y^2 - 4y - 4 + \lambda(x^2 + 2x + y^2 + 4y - 4) = 0 $

This simplifies to:

$ x^2(1+\lambda) + y^2(1+\lambda) + x(-2 + 2\lambda) + y(-4 + 4\lambda) - 4 - 4\lambda = 0 \tag{i} $

The circle in equation (i) must pass through the point $(3,3)$, so we substitute these coordinates into the equation:

$ \begin{aligned} &9(1 + \lambda) + 9(1 + \lambda) + 3(-2 + 2\lambda) + 3(-4 + 4\lambda) - 4 - 4\lambda = 0 \end{aligned} $

Simplifying the above expression yields:

$ 18 + 18\lambda - 6 + 6\lambda - 12 + 12\lambda - 4 - 4\lambda = 0 $

This further simplifies to:

$ 32\lambda - 4 = 0 \implies \lambda = \frac{1}{8} $

Substituting $\lambda = \frac{1}{8}$ back into equation (i), the equation of the circle becomes:

$ \begin{aligned} &\frac{9}{8}x^2 + \frac{9}{8}y^2 - \frac{14}{8}x - \frac{28}{8}y - \frac{36}{8} = 0 \\ \Rightarrow &x^2 + y^2 - \frac{14}{9}x - \frac{28}{9}y - 4 = 0 \end{aligned} $

From this, we identify $\alpha = \frac{-14}{9}$, $\beta = \frac{-28}{9}$, and $\gamma = -4$.

Now, calculating $3(\alpha + \beta + \gamma)$:

$ 3\left(-\frac{14}{9} - \frac{28}{9} - 4\right) = -\frac{78}{3} = -26 $

Given circle

$ \begin{aligned} & x^2+y^2-2 x+4 y+4=0 \\ & \Rightarrow(x-1)^2+(y+2)^2=1 \end{aligned} $

Circle is centred at $O(1,-2)$ and $r=1$

Given chord having equation $x+y=1$

Let us find the intersection points of circle and chord.

$ \begin{aligned} & \\ & \Rightarrow \quad y^2+y^2+4 y+4=1 \\ & \Rightarrow \quad \quad 2 y^2+4 y+3=0 \\ & \because a>0 \text { and } D<0 \end{aligned} $

Hence, $x+y-1$ is not the chord of the circle $x^2+y^2-2 x+4 y+4=0$

Given a point $ P $ on the circle $ x^2 + y^2 = 25 $, let the coordinates of $ P $ be $(x_1, y_1)$. Thus, we have:

$ x_1^2 + y_1^2 = 25 \quad \text{(Equation 1)} $

The chord of contact from point $ P $ to the circle $ x^2 + y^2 = 9 $ is given by the equation:

$ xx_1 + yy_1 = 9 $

Let $(h, k)$ be the pole of this chord with respect to the circle $ x^2 + y^2 = 36 $. The polar of $ (h, k) $ is given by:

$ hx + ky = 36 $

Since the line $ L $ is the polar of $ P $ with respect to the circle $ x^2 + y^2 = 36 $, the equation of the polar line, in terms of $ (x_1, y_1) $, is simply:

$ xx_1 + yy_1 = 0 $

The ratio given by comparing the polar equations is:

$ \frac{h}{x_1} = \frac{k}{y_1} = \frac{36}{9} = 4 $

This implies:

$ x_1 = \frac{h}{4} \quad \text{and} \quad y_1 = \frac{k}{4} $

Substituting $ x_1 $ and $ y_1 $ from above into Equation 1, we have:

$ \left(\frac{h}{4}\right)^2 + \left(\frac{k}{4}\right)^2 = 25 $

Simplifying:

$ \frac{h^2}{16} + \frac{k^2}{16} = 25 $

Multiplying through by 16:

$ h^2 + k^2 = 400 $

Therefore, the locus of the poles $(h, k)$ is given by:

$ x^2 + y^2 = 400 $

We are given the circle equation:

$ S \equiv x^2 + y^2 - 14x + 6y + 33 = 0 $

The center, $ C_1 $, of this circle is $ (7, -3) $, and we can determine its radius, $ r_1 $, as follows:

$ r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49 + 9 - 33} = \sqrt{25} = 5 $

For the circle $ S^1 \equiv x^2 + y^2 = a^2 $, the center, $ C_2 $, is $ (0, 0) $ and its radius, $ r_2 $, is $ a $.

To have four common tangents between the two circles, the condition is:

$ C_1C_2 > r_1 + r_2 $

Calculating $ C_1C_2 $, we have:

$ C_1C_2 = \sqrt{(7 - 0)^2 + (-3 - 0)^2} = \sqrt{49 + 9} = \sqrt{58} $

Thus, we require:

$ \sqrt{58} > a + 5 $

Approximating $ \sqrt{58} \approx 7.6 $, we get:

$ a + 5 < 7.6 $

$ a < 7.6 - 5 $

$ a < 2.6 $

Since $ a $ is a natural number ($ a \in \mathbb{N} $), the possible values of $ a $ are 1 and 2.

Therefore, there are 2 possible values of $ a $.

The line $2x + 5y + \alpha = 0$ intersects the $x$-axis and $y$-axis at points $A\left(-\frac{\alpha}{2}, 0\right)$ and $B\left(0, -\frac{\alpha}{5}\right)$ respectively. This forms a right-angled triangle with the axes.

The circumcenter of a right-angled triangle is at the midpoint of the hypotenuse $AB$. Thus, the center of the circumcircle is $\left(\frac{-\alpha}{4}, \frac{-\alpha}{10}\right)$.

To find the radius of the circumcircle, we calculate the distance from the center to one of the vertices, say $A$:

$ \begin{aligned} \text{Radius of the circle} &= \sqrt{\left(\frac{-\alpha}{2} - \frac{-\alpha}{4}\right)^2 + \left(0 - \frac{-\alpha}{10}\right)^2} \\ &= \sqrt{\left(-\frac{\alpha}{4}\right)^2 + \left(\frac{\alpha}{10}\right)^2} \\ &= \sqrt{\frac{\alpha^2}{16} + \frac{\alpha^2}{100}} \\ &= \sqrt{\frac{25\alpha^2 + 4\alpha^2}{400}} \\ &= \frac{\sqrt{29}\alpha}{20}. \end{aligned} $

The area of the circumcircle is given by $\pi r^2$:

$ \begin{aligned} \pi \left(\frac{29}{400} \alpha^2\right) &= \frac{29\pi}{4} \\ \Rightarrow \quad \frac{29\alpha^2}{400} &= \frac{29}{4} \\ \Rightarrow \quad \alpha^2 &= 100 \\ \Rightarrow \quad |\alpha| &= 10. \end{aligned} $

We are given a circle defined by the equation:

$ S \equiv x^2 + y^2 - 2x - 4y + 1 = 0 $

This can be rewritten by completing the square as:

$ (x - 1)^2 + (y - 2)^2 = 4 $

To find where this circle intersects the Y-axis, set $ x = 0 $:

$ y^2 - 4y + 1 = 0 $

Solving this quadratic equation:

$ y = \frac{4 \pm \sqrt{16 - 4}}{2} $

$ y = 2 \pm \sqrt{3} $

Thus, the points of intersection are $ A(0, 2+\sqrt{3}) $ and $ B(0, 2-\sqrt{3}) $. Since $ OA > OB $, we have $ OA = 2 + \sqrt{3} $ and $ OB = 2 - \sqrt{3} $.

The radical axis of the circles $ S \equiv x^2 + y^2 - 2x - 4y + 1 = 0 $ and $ S' \equiv x^2 + y^2 - 4x - 2y + 4 = 0 $ is found by subtracting:

$ S - S' = 0 $

$ (x^2 + y^2 - 2x - 4y + 1) - (x^2 + y^2 - 4x - 2y + 4) = 0 $

$ -2x + 4x - 4y + 2y + 1 - 4 = 0 $

$ 2x - 2y - 3 = 0 $

This simplifies to the line equation:

$ 2x - 2y = 3 $

Now, solve for $ y $ when $ x = 0 $ (where it intersects the Y-axis):

$ 2(0) - 2y = 3 $

$ y = -\frac{3}{2} $

Thus, the point $ C $ where it intersects the Y-axis is $ (0, -\frac{3}{2}) $.

Assume $ C $ divides $ AB $ in the ratio $ k:1 $. Using section formula for coordinates:

$ -\frac{3}{2} = \frac{k(2 - \sqrt{3}) + (2 + \sqrt{3})}{k + 1} $

Solving for $ k $:

$ -3k - 3 = 4k - 2\sqrt{3}k + 4 + 2\sqrt{3} $

Simplifying:

$ -7k + 2\sqrt{3}k = 7 + 2\sqrt{3} $

$ k = \frac{7 + 2\sqrt{3}}{-7 + 2\sqrt{3}} $

Therefore, the ratio in which $ C $ divides the segment $ AB $ is:

$ (7 + 2\sqrt{3}) : (-7 + 2\sqrt{3}) $