Circle

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}) $

Let the equation of the desired circle be:

$ S \equiv x^2 + y^2 + 2gx + 2fy + c = 0 $

The circle $ S $ is orthogonal to three given circles. Using the property of orthogonal circles, we find the equations involving $ g, f, $ and $ c $.

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

Since $ S $ and this circle cut orthogonally, we have:

$ 2 \cdot g \cdot (-1) + 2 \cdot f \cdot 3 = c + 0 $

Simplifying:

$ -2g + 6f = c \quad \text{...(i)} $

For the Circle $ x^2 + y^2 - 4x - 2y + 6 = 0 $:

Orthogonality gives us:

$ 2g \cdot (-2) + 2f \cdot (-1) = c + 6 $

Simplifying:

$ -4g - 2f = c + 6 $

For the Circle $ x^2 + y^2 - 12x + 2y + 3 = 0 $:

Orthogonality yields:

$ 2g \cdot (-6) + 2f \cdot 1 = c + 3 $

Simplifying:

$ -12g + 2f = c + 3 $

From equation (i), substituting $ c $ from the orthogonality conditions, we proceed to simplify:

Solving the System of Equations:

Using equations:

$ -4g - 2f = c + 6 $

$ -12g + 2f = c + 3 $

Rearrange equation (i):

$ c = -2g + 6f \quad \text{(equation i)} $

From $-4g - 2f = -2g + 6f + 6$:

$ -2g - 8f = 6 \quad \Rightarrow g + 4f = -3 \quad \text{...(ii)} $

From $-12g + 2f = -2g + 6f + 3$:

$ -10g - 4f = 3 \quad \Rightarrow 10g + 4f = -3 $

Solving equations (ii) and the above, we find:

$ g = 0, \quad f = -\frac{3}{4} $

Substituting into equation (i) for $ c $:

$ c = 6 \times \left(-\frac{3}{4}\right) = -\frac{9}{2} $

Equation of the Circle $ S $:

$ x^2 + y^2 - \frac{3}{2}y - \frac{9}{2} = 0 $

Finding the Equation of the Tangent at $ (0, 3) $:

The equation of the tangent to this circle at the point $(x_1, y_1)$ is given by:

$ x x_1 + y y_1 - \frac{3}{2} \left(\frac{y + y_1}{2}\right) - \frac{9}{2} = 0 $

Substituting point $ (0, 3) $:

$ 3y - \frac{3}{2} \left(\frac{y + 3}{2}\right) - \frac{9}{2} = 0 $

Simplifying:

$ 3y - \frac{3}{4}y - \frac{9}{4} - \frac{9}{2} = 0 $

Convert all terms to common denominator:

$ \frac{12y}{4} - \frac{3y}{4} - \frac{9}{4} - \frac{18}{4} = 0 $

Re-arrange the equation:

$ \frac{9y}{4} = \frac{27}{4} $

Solve for $ y $:

$ y = 3 $

Thus, the equation of the tangent is:

$ y = 3 $

We have,

Equation, of circle is

$ x^2+y^2-6 x+4 y+12=0 $

Centre of circle $=(\lambda-2)$

radius $=\sqrt{9+4-12}=1$

$ \begin{aligned} O P & =\sqrt{(3-2)^2+(-2-3)^2} \\ & =\sqrt{1+25}=\sqrt{26} \\ A P & =\sqrt{O P^2-A O^2} \\ & =\sqrt{26-1}=5 \end{aligned} $

$\ln \triangle M O P, \tan \frac{\theta}{2}=\frac{A O}{A P}=\frac{1}{5}$

$ \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^2}=\frac{\frac{2}{5}}{\frac{24}{25}}=\frac{5}{12} $

Hence, $\theta=\tan ^{-1} \frac{5}{12}$

Let ( $a, b$ be the point on the line

$ x+2 y+K=0 $

We have, equation of circle is

$ x^2+y^2+8 x-6 y-24=0 $

Equation of polar for the given circle is obtained as

$ \begin{aligned} & a x+b y+4(x+d)-3(y+b)-24=0 \\ & (a+4 x+(b-3) y+4 a-3-24 \end{aligned} $

Now, comparing this line with

$2 x-3 y+3=0$ we get

$ \begin{aligned} & \frac{a+4}{2}=\frac{b-3}{-3}=\frac{4 a-y-24}{3} \\ & -3 a-12=2 b-6 \Rightarrow 3 a+2 b=-6 \\ & \text { and } 3 a+12=8 a-6 b-4 s \\ & \Rightarrow 5 a-6 b=60 \ldots \text { (ii) } \end{aligned} $

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

$ a=3 \text { and } b=\frac{-15}{2} $

Now. $\left(3, \frac{15}{2}\right)$ lies on line $x+2 y+k=0$

$ 3-15+K=0 \Rightarrow K=12 $

Required length of tangent from a point

$ \begin{aligned} &\left(\frac{12}{4}, \frac{12}{3}\right) \text { i.e. }(3,4) \\ &=\sqrt{3^3+4^2+24-24-24} \\ &=\sqrt{9+16-24}=1 \end{aligned} $

The equation of pole with respect ts the point $\beta, 2$ to the given circle is

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

Inverse of point $P(1,2)$ is the foot $(h, k)$ of the perpendicalar from the point $(1,2)$

$ \text { to the line } x-2 y+1=0 $

$ \begin{array}{ll} \Rightarrow & \frac{h-1}{1}=\frac{k-2}{-2}=-\left[\frac{1-4+1}{1^2+\left(-2^3\right.}\right] \\ \Rightarrow & h-1=\frac{K-2}{-2}=\frac{2}{5} \\ \Rightarrow & h-1=\frac{2}{5} \text { and } \frac{K-2}{-2}=\frac{2}{5} \\ \Rightarrow & h=\frac{7}{5} \text { and } K=\frac{-4}{5}+2=\frac{6}{5} \end{array} $

Hence, $2 h+K=\frac{14}{5}+\frac{6}{5}=\frac{20}{5}=4$

We have, equation of two circlet

$ x^2+y^2+4 x-5=0 $

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

Centre and radius of both circles are

$ \begin{aligned} & c_1=\left(-2,0, r_1=\sqrt{4+5}=3\right. \\ & c_2=(0,3), r_2=\sqrt{9-8}=1 \end{aligned} $

Let $P$ and $Q$ be the internal and external centre of similitude, it divides $C_1 C_2$ is ine ratio 3 : 1.

$ P=\frac{x_2+C_1}{3+1}=\left(\frac{-2}{4}, \frac{9}{4}\right)=\left(\frac{-1}{2} ; \frac{9}{4}\right) $

Here, $a=\frac{-1}{2}$ and $b=\frac{9}{4}$

$ Q=\frac{3 C_2-C_1}{C_1-C_2}=\left(\frac{2}{2}, \frac{9}{2}\right)=\left(1, \frac{9}{2}\right) $

Here, $C=1$ and $d=\frac{9}{2}$

$ \begin{aligned} \text { Hence, }(a+\Delta)(b+c) & =\left(\frac{-1}{2}+\frac{9}{2}\right)\left(\frac{9}{4}+1\right) \\ & =\frac{4 \times 13}{4}=13 \end{aligned} $

The equation of a circle that passes through the intersection points of the given circles is determined as follows:

First, start with the general form that includes both circles and a parameter $\lambda$:

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

Simplifying this equation leads to:

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

To rewrite in standard circle form, factor out the coefficients:

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

The center of the circle $(h, k)$ from this equation is:

$ \left(\frac{1 - \lambda}{1 + \lambda}, \frac{\lambda - 1}{1 + \lambda}\right) $

Since this center lies on the line $x - y + 6 = 0$, we substitute the center into the line's equation:

$ 1 - \lambda - \left(\lambda - 1\right) + 6(1 + \lambda) = 0 $

Solving:

$ 1 - \lambda - \lambda + 1 + 6\lambda + 6 = 0 \quad \Rightarrow \quad 4\lambda + 8 = 0 \quad \Rightarrow \quad \lambda = -2 $

Substitute $\lambda = -2$ back into the center, we get:

$ (h, k) = \left(-3, 3\right) $

Finally, calculate the radius of the circle:

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

Solving further:

$ \text{Radius} = \sqrt{9 + 9 + 4} = \sqrt{14} $

Given $ x=a\left(\frac{1-t^{2}}{1+t^{2}}\right) $ and $ y=\frac{2 a t}{1+t^{2}} $

Let $ \quad t=\tan \theta $

$ \Rightarrow \quad x=a\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) $

and $ y=\frac{2 a \tan \theta}{1+\tan ^{2} \theta} $

$ \Rightarrow \quad x=a \cos 2 \theta $ and $ y=a \sin 2 \theta $

$ \Rightarrow \cos 2 \theta=\frac{x}{a} $ and $ \sin 2 \theta=\frac{y}{a} $

Squaring both sides, we get

$ \cos ^{2} 2 \theta=\frac{x^{2}}{a^{2}} \quad \ldots \text { (i) } $

and $ \sin ^{2} 2 \theta=\frac{y^{2}}{a^{2}} \quad \ldots \text { (ii) } $

Adding Eqs. (i) and (ii), we get

$ \cos ^{2} 2 \theta+\sin ^{2} 2 \theta=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}} $

$ \Rightarrow \quad x^{2}+y^{2}=a^{2} $

$ \therefore $ Locus is a circle having centre at origin and radius ' $ a $ '.

According to question,

$ O C=\sqrt{h^{2}+k^{2}} $

In $ \triangle O C B $,

$ \cos 45^{\circ}=\frac{\sqrt{h^{2}+k^{2}}}{2} $

$ \Rightarrow \quad \frac{\sqrt{h^{2}+k^{2}}}{2}=\frac{1}{\sqrt{2}} $

$ \Rightarrow \quad h^{2}+k^{2}=2 $

Replacing $ h \rightarrow x $ and $ k \rightarrow y $

Locus $ \Rightarrow x^{2}+y^{2}=2 $

We are given two circles:

(1) $x^2+y^2-18 x-15 y+131=0$

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

First, let's find the centers and radii of the circles.

For circle (1):

Completing the square for the equation:

$(x^2-18x+{81})+(y^2-15y+\frac{225}{4})=-131+{81}+\frac{225}{4}$

$(x-9)^2+(y-\frac{15}{2})^2=\frac{25}{4}$

Center 1: $C_1(9, \frac{15}{2})$

Radius 1: $r_1 = \sqrt{\frac{25}{4}}=\frac{5}{2}$

For circle (2):

Completing the square for the equation:

$(x^2-6x+9)+(y^2-6y+9)=7+9+9$

$(x-3)^2+(y-3)^2=25$

Center 2: $C_2(3, 3)$

Radius 2: $r_2 = 5$

Now, let's find the distance between the centers :

$d = \sqrt{(9-3)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + \frac{9}{2}^2} = \sqrt{36 + \frac{81}{4}} = \frac{15}{2}$

Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii :

1. If $d > r_1 + r_2$, the circles are separate, and there are 4 common tangents.
2. If $d = r_1 + r_2$, the circles are externally tangent, and there are 3 common tangents.
3. If $0 < d < |r_1 - r_2|$, one circle is inside the other, and there are no common tangents.
4. If $d = |r_1 - r_2|$, the circles are internally tangent, and there is 1 common tangent.
5. If $d < |r_1 - r_2|$, one circle is completely inside the other, and there are no common tangents.

In this case :

$d = \frac{15}{2}$

$r_1 = \frac{5}{2}$

$r_2 = 5$

Now, let's check the conditions :

$r_1 + r_2 = \frac{5}{2} + 5$ = $\frac{15}{2}$

Since $d = \frac{15}{2} = r_1 + r_2$, the circles touch each other externally, and there are 3 common tangents.

$(h,k) = \left( {{{5.5 + 5( - 2) + 14\sqrt 2 } \over {10 + 7\sqrt 2 }},{{35 + 21\sqrt 2 } \over {10 + 7\sqrt 2 }}} )\right.$

$h + k = {{50 + 35\sqrt 2 } \over {10 + 7\sqrt 2 }} = 5$

Equation of AB :

$y - ( - 5) = {{ - 5 + 11} \over {8 - 4}}(x - 8)$

$ \Rightarrow y + 5 = {3 \over 2}(x - 8)$

$ \Rightarrow 2y + 10 = 3x - 24$

$ \Rightarrow 3x - 2y - 34 = 0$ .... (1)

Also AB is cord of contact. And we know equation of cord of contact to a circle is $T = 0$

$ \Rightarrow xh + yk - {3 \over 2}(x + h) + 5(y + k) - 15 = 0$

$ \Rightarrow x\left( {h - {3 \over 2}} \right) + y(k + 5) + \left( { - {3 \over 2}h + 5k - 15} \right) = 0$ .... (2)

Comparing equation (1) and (2), we get

${{h - {3 \over 2}} \over 3} = {{k + 5} \over { - 2}} = {{ - {3 \over 2}h + 5k - 15} \over { - 34}}$

$\therefore$ $ - 2h + 3 = 3k + 15$

$ \Rightarrow 3k + 2h = - 12$ ..... (3)

and

$17(k + 5) = - {3 \over 2}h + 5k - 15$

$ \Rightarrow 12k + {3 \over 2}h - 100$

$ \Rightarrow 3h + 24k = - 200$ ..... (4)

Solving (3) and (4), we get

$h = 8$ and $k = - {{28} \over 3}$

$\therefore$ Point C is $\left( {8, - {{28} \over 3}} \right)$

Now radius of the circle whose centre is at C and tangent is AB is

$ = \left| {{{3(8) - 2\left( { - {{28} \over 3}} \right) - 34} \over {\sqrt {{3^2} + {2^2}} }}} \right|$

$ = \left| {{{26} \over {3\sqrt {13} }}} \right|$

$ = {{2\sqrt {13} } \over 3}$

$\therefore \cos \left(\frac{\theta_{1}}{2}\right)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{\theta_{i}}{2}\right)$

Given $r_{1}^{2}=r_{2}^{2}+r_{3}^{3}$

$\Rightarrow\left(\cos \left(\frac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\frac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\frac{\theta_{3}}{2}\right)\right)^{2}$

$\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{\theta_{2}}{2}\right)+\frac{1}{4}$

$\Rightarrow \cos ^{2}\left(\frac{\theta_{2}}{2}\right)=\frac{1}{2}$

$\Rightarrow \frac{\theta_{2}}{2}=\frac{\pi}{4}$

$\Rightarrow \theta_{2}=\frac{\pi}{2}$

$A(a, 3), B(b, 5), C(a, b)$ circumcentre $O(1,1)$

We know that, $O A^2=O B^2=O C^2$

On equating II and III, we get

$ \begin{aligned} & (a-1)^2=16 \Rightarrow a-1= \pm 4 \\ & \Rightarrow \quad a=5,-3 \end{aligned} $

On equating I and III, we get

$ (b-1)^2=4 \Rightarrow b-1= \pm 2 \Rightarrow b=3,-1 $

Sum of absolute distinct values

$ =5+3+1=9 $

$ \begin{aligned} &\text { Passes through }(-6,3)\\ &\begin{aligned} & \text { Radius }=a \\ & c(-a, a) \end{aligned} \end{aligned} $

Equation will be

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

At $(-6,3)$,

$ \begin{aligned} 36+9-12 a-6 a+a^2 & =0 \\ \Rightarrow \quad a^2-18 a+45 & =0 \end{aligned} $

$⇒ a=15,3 $

Equation of circle will be

$ \begin{aligned} & x^2+y^2+30 x-30 y+225=0 \\ & x^2+y^2+6 x-6 y+9=0 \end{aligned} $