Circle

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} $

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} $

$x^2+y^2-10 x=0$

Tangent at $(9,3)$

$ \begin{aligned} & T=0 \Rightarrow 9 x+3 y-5(x+9)=0 \\ & -4 x+3 y=45 \end{aligned} $

$ \begin{aligned} &\text { Normal at }(9,3)\\ &\begin{aligned} & 3 x-4 y=\lambda \Rightarrow 3 \times 9-4 \times 3=\lambda \\ & \Rightarrow \lambda=15 \Rightarrow 3 x-4 y=15 \\ & \text { Area of triangle }=\frac{1}{2} \times\left(\frac{45}{4}-5\right) \times 3=\frac{75}{8} \end{aligned} \end{aligned} $

Given, circles

$ \begin{aligned} & x^2+y^2-4=0 \text { and } \\ & (x-3)^2+(y-4)^2=49 \\ & c_1\left(0,0, r_1=2 \text { and } c_2(3,4), r_2=7\right. \\ & \text { Now, } c_1 c_2=\sqrt{9+16}=5 \\ & \left|r_1-r_2\right|=5 \\ & c_1 c_2=\left|r_1-r_2\right| \end{aligned} $

⇒ Circles touch internally.

Given, circle

$ x^2+y^2+2 x+8 y-23=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of new circle touching

Eq. (i) at point $(1,2)$ is

$ \begin{gathered} x^2+y^2+2 x+8 y-23+\lambda(x-1)^2 \left.+(y-2)^2\right]=0 \\ \Rightarrow(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(8-4 \lambda) y-23+5 \lambda=0 \\ \Rightarrow x^2+y^2-2 x\left(\frac{\lambda-1}{\lambda+1}\right)-4 y\left(\frac{\lambda-2}{\lambda+1}\right) +\frac{5 \lambda-23}{\lambda+1}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \text { } \end{gathered} $

$ \begin{aligned} &\text { Given that } r=\sqrt{10}\\ &\begin{aligned} & \Rightarrow\left(\frac{\lambda-1}{\lambda+1}\right)^2+4\left(\frac{\lambda-2}{\lambda+1}\right)^2-\frac{5 \lambda-23}{\lambda+1}=10 \\ & \Rightarrow \lambda^2-2 \lambda+1+4 \lambda^2-16 \lambda+16 -\left(5 \lambda^2-18 \lambda-23\right)=10(\lambda+1)^2 \\ & \Rightarrow \quad 10 \lambda^2+20 \lambda-30=0 \\ & \Rightarrow \quad \lambda^2+2 \lambda-3=0 \end{aligned} \end{aligned} $

$ \Rightarrow \quad \lambda=-3,1 $

At $\lambda=1 \Rightarrow$ Eq. (ii)becomes

$ \begin{array}{lr} \Rightarrow & x^2+y^2+2 y-9=0 \\ \Rightarrow & |a+b+c|=7 \end{array} $

At $\lambda=-3 \Rightarrow$ Eq. (ii) becomes

$ \begin{aligned} & \Rightarrow x^2+y^2-4 x-10 y+19=0 \\ & \Rightarrow \quad|a+b+c|=5 \end{aligned} $

$x^2+y^2-4 x-6 y+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let the circle passing through origin centred on line $x-y=0$

$ x^2+y^2+2 g x+2 f y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∵ Line (i) and circle (ii) cuts orthogonally

$ \begin{aligned} & \left\{\begin{array}{l} g=f \\ \text { and } c=0 \end{array}\right. \\ & -4 g-6 g=0+10 \Rightarrow g=-1 \end{aligned} $

Now, radius of (ii) circle

$ r=\sqrt{g^2+f^2-c}=\sqrt{2} $

Diameter $=2 r=2 \sqrt{2}$

$ \begin{aligned} &\text { Equation of required circle is }\\ &\begin{aligned} & \begin{array}{l} x^2+y^2+6 x+4 y-12+\lambda \\ \quad\left(x^2+y^2-4 x-6 y-12\right)=0 \end{array} \\ & \Rightarrow x^2+y^2+\frac{(6-4 \lambda) x}{\lambda+1}+\frac{(4-6 \lambda) y}{\lambda+1} -\frac{12(\lambda+1)}{\lambda+1}=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Radius }=\sqrt{13} \\ & \begin{array}{l} \Rightarrow \sqrt{\frac{(2 \lambda-3)^2}{(\lambda+1)^2}+\frac{(3 \lambda-2)^2}{(\lambda+1)^2}+12}=\sqrt{13} \\ \Rightarrow \quad(\lambda+1)^2=13 \lambda^2+13-24 \lambda \\ \Rightarrow 12 \lambda^2-26 \lambda+12=0 \\ \Rightarrow \quad 6 \lambda^2-13 \lambda+6=0 \\ \quad \lambda=\frac{3}{2}, \frac{2}{3} \end{array} \\ & \text { At } \lambda=\frac{3}{2} \Rightarrow x^2+y^2-2 y-12=0 \\ & \text { At } \lambda=\frac{2}{3} \Rightarrow x^2+y^2+2 x-12=0 \end{aligned} $

Let, coordinates of $P=(x, y)$

So, distance of $P$ from

$ (0,2)=\sqrt{(x-0)^2+(y-2)^2} $

and distance of $P$ from

$ (-1,0)=\sqrt{(x+1)^2+(y-0)^2} $

According to the question,

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

On squaring both sides, we get

$ \begin{aligned} & 2\left(x^2+y^2-4 y+4\right)=x^2+y^2+2 x+1 \\ & \Rightarrow x^2+y^2-2 x-8 y+7=0 \end{aligned} $

which is the equation of a circle.

$ \begin{aligned} \text { Centre of circle } & =\left(\frac{-(-2)}{2}, \frac{-(-8)}{2}\right) \\ & =(1,4) \end{aligned} $

and radius $(r)=\sqrt{(1)^2+(4)^2-7}=\sqrt{10}$

Hence, a circle with centre $(1,4)$ and radius $\sqrt{10}$ units.

$(3,4),(3,2)$ and $(1,4)$

$ \begin{gathered} x=a+r \cos \theta, y=b+r \sin \theta, \\ \cos \theta=\frac{x-a}{r}, \sin \theta=\frac{y-b}{r} \\ \because \quad \sin ^2 \theta+\cos ^2 \theta=1 \\ \left(\frac{x-a}{r}\right)^2+\left(\frac{y-b}{r}\right)^2=1 \end{gathered} $

Equation of circle $(x-a)^2+(y-b)^2=r^2$

$ x^2+y^2-2 a x-2 b y+a^2+b^2=r^2 $

Putting the point $(3,4),(3,2),(1,4)$ respectively.

$ \begin{aligned} & (3,4) \rightarrow 9+16-6 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-8 b+25=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & (3,2) \rightarrow 9+4-6 a-4 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-4 b+13=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & (1,4) \rightarrow 1+16-2 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-2 a-8 b+17=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtracting Eq. (ii) from Eq. (i),

$ \begin{array}{ll} & -4 b+12=0 \\ \Rightarrow & b=3 \end{array} $

Subtracting Eq. (iii) from Eq. (i),

$ \begin{array}{r} -4 a+8=0 \\\Rightarrow \,\,\,\,\,\,\,\,\, a=2 \end{array} $

Putting values of $a$ and $b$ in Eq. (i),

$ \begin{array}{lc} & 4+9-12-24+25=r^2 \\ \Rightarrow & r^2=2 \Rightarrow r=\sqrt{2} \\ \therefore & b^a r^a=3^2(\sqrt{2})^2=9 \times 2=18 \end{array} $

Given, equation of circle is

$ x^2+y^2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiating Eq. (i) w.r.t. $x$ on both sides, we get

$ \begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ \Rightarrow \quad & 2 y \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=\frac{-x}{y} \end{aligned} $

∴ Slope of tangent to $x^2+y^2=4$ at

$P(\sqrt{3}, 1)$ is given by $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}$.

So, $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$

So, slope of line $L=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$

$ \left[\because \text { slope of line }=\frac{-1}{d y / d x}\right] $

Let equation of line $L$ be $x-\sqrt{3} y+c=0$

As, $L$ is tangent to $(x-3)^2+y^2=1$.

So, perpendicular distance of $L$ from its centre should be equal to its radius i.e. 1. Centre $(3,0)$

$ \therefore \quad \frac{|3+c|}{\sqrt{4}}=1 \Rightarrow c=-1 \text { or }-5 $

So, equation of line $L$ is $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$

$x^2+y^2-2 x+4 y+3=0$ point

$ \begin{aligned} & P(6,-5) \\ & (x-1)^2+(y+2)^2=2 \end{aligned} $

Center $(1,-2)$

Radius $=\sqrt{2}$

Equation of tangent to given circle

$ \begin{aligned} y+2 & =m(x-1) \pm \sqrt{2} \sqrt{1+m^2} \\ (\because y-k & \left.=m(x-h) \pm a \sqrt{1+m^2}\right) \end{aligned} $

$m$ is slope of the tangent.

Tangent is passing through $(6,-5)$

$ \begin{aligned} -5+2 & =m(6-1) \pm \sqrt{2} \sqrt{1+m^2} \\ -3 & =5 m \pm \sqrt{2} \sqrt{1+m^2} \\ -3-5 m & = \pm \sqrt{2} \sqrt{1+m^2} \end{aligned} $

Squaring on both sides,

$ \begin{array}{rr} & 9+25 m^2+30 m=2\left(1+m^2\right) \\ \Rightarrow & 23 m^2+30 m+7=0 \\ \Rightarrow & 23 m^2+23 m+7 m+7=0 \\ \Rightarrow & (23 m+7)(m+1)=0 \\ \Rightarrow & m_1=-1, m_2=-\frac{7}{23} \end{array} $

$ \begin{array}{ll} & \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-1+\frac{7}{23}}{1+\frac{7}{23}}\right| \\ \Rightarrow & \tan \theta=\left|\frac{-16}{30}\right| \Rightarrow \tan \theta=\frac{16}{30}=\frac{8}{15} \\ \Rightarrow & \tan \theta=\frac{8}{15} \Rightarrow \cot \theta=\frac{15}{8} \end{array} $

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

Centre $\left(c_1\right)=(2,3)$

and $r_1=\sqrt{(2)^2+(3)^2-k}=\sqrt{13-k}$

and $x^2+y^2+8 x-4 y+11=0$

Centre $\left(c_2\right)=(-4,2)$

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

Now,

$ d=c_1 c_2=\sqrt{(2+4)^2+(3-2)^2}=\sqrt{37} $

As, we know

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

$ \begin{aligned} & \Rightarrow \cos \pi / 2=\frac{13-k+9-37}{2(\sqrt{13-k}) \times 3} \\ & \Rightarrow \quad 0=22-37-k \Rightarrow k=-15 \end{aligned} $

$(x \pm \lambda)^2+(y \pm \lambda)^2=\lambda^2$

Centres of four circles are $\left(\lambda_{,}{ }^A \lambda\right),\left(-\lambda_{,}^B, \lambda\right),\left(-\lambda_{,}{ }^C-\lambda\right)$ and $(\lambda,-\lambda)$. Radius of circles are $\lambda$.

Radius of possible circles are, $O P$ or $O Q$

$ \begin{aligned} O Q & =O A+A Q=O A+\lambda \\ \text { and } O A & =\sqrt{(\lambda-0)^2+(\lambda-0)^2} \\ & =\lambda \sqrt{2} \end{aligned} $

So, $O Q=\lambda(\sqrt{2}+1)$

and $O P=\lambda \sqrt{2}-\lambda=\lambda(\sqrt{2}-1)$

From the given options

Then, radius $=(\sqrt{2}-1) \lambda$

$ \begin{aligned} & \text { } S_1: x^2+y^2-1=0 \\ & S_2: x^2+y^2-2 x-3=0 \\ & S_3: x^2+y^2-2 y-3=0 \end{aligned} $

For radical circle,

$ \begin{aligned} & S_1-S_2=0 \\ & \Rightarrow-1+2 x+3=0 \Rightarrow x=-1 \\ & \text { and } S_2-S_3=0 \Rightarrow-2 x-3+2 y+3=0 \\ & \Rightarrow x=y=-1 \end{aligned} $

So, centre of radical circle $=C(\alpha, \beta)$

$ =(-1,-1) $

and $r_1=1, r_2=\sqrt{(1)^2-(-3)}=2$

and $r_3=\sqrt{(+1)^2-(-3)}=2$

So, $r=r_1+r_2+r_3=5$

Therefore, equation of circle

$ (x+1)^2+(y+1)^2=25 $

Equation of $H F$, which is mid-parallel to $A D$ and $B C$

$ x+y-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of $G E$, which is mid-parallel to $A B$ and $C D$

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

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

$ x=(1 / 2) \text { and } y=(7 / 2) $

∴ Centre $=(1 / 2,7 / 2)$

$ \text { and radius }=\frac{1}{2} \times \frac{|-6-(-2)|}{\sqrt{1^2+1^2}}=\frac{2 \sqrt{2}}{2}=\sqrt{2} $

∴ Equation of circle

$ \begin{aligned} \left(x-\frac{1}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=(\sqrt{2})^2 \\ \Rightarrow \quad 2 x^2+2 y^2-2 x-14 y+21 & =0 \end{aligned} $