Circle

Center of given circle is

Radius

Equation of AB :

.... (1) Also AB is cord of contact. And we know equation of cord of contact to a circle is

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

$\therefore$

..... (3) and

..... (4) Solving (3) and (4), we get

and

$\therefore$ Point C is

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

As $A$ and $B$ satisfy both line and circle we have $\alpha=0 \Rightarrow A(0,0)$ and $\beta=1$ i.e.

$B(1,1)$ Centre of circle as $A B$ diameter is $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ and radius $=\dfrac{1}{\sqrt{2}}$ $\therefore$ For image of $\left(\dfrac{1}{2} ; \dfrac{1}{2}\right)$ in $x+y+z$ we get $\dfrac{x-\dfrac{1}{2}}{1}=\dfrac{y-\dfrac{1}{2}}{1}=\dfrac{-2(3)}{2}$ $\Rightarrow$ Image $\left(-\dfrac{5}{2},-\dfrac{5}{2}\right)$ $\therefore$ Equation of required circle $\left(x+\dfrac{5}{2}\right)^{2}+\left(y+\dfrac{5}{2}\right)^{2}=\dfrac{1}{2}$ $\Rightarrow x^{2}+y^{2}+5 x+5 y+\dfrac{50}{4}-\dfrac{1}{2}=0$ $\Rightarrow x^{2}+y^{2}+5 x+5 y+12=0$

$\therefore \cos \left(\dfrac{\theta_{1}}{2}\right)=\dfrac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\dfrac{\theta_{i}}{2}\right)$ Given $r_{1}^{2}=r_{2}^{2}+r_{3}^{3}$ $\Rightarrow\left(\cos \left(\dfrac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\dfrac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\dfrac{\theta_{3}}{2}\right)\right)^{2}$ $\Rightarrow \dfrac{3}{4}=\cos ^{2}\left(\dfrac{\theta_{2}}{2}\right)+\dfrac{1}{4}$ $\Rightarrow \cos ^{2}\left(\dfrac{\theta_{2}}{2}\right)=\dfrac{1}{2}$ $\Rightarrow \dfrac{\theta_{2}}{2}=\dfrac{\pi}{4}$ $\Rightarrow \theta_{2}=\dfrac{\pi}{2}$

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: $(x^2-18x+{81})+(y^2-15y+\dfrac{225}{4})=-131+{81}+\dfrac{225}{4}$ $(x-9)^2+(y-\dfrac{15}{2})^2=\dfrac{25}{4}$ Center 1: $C_1(9, \dfrac{15}{2})$ Radius 1: $r_1 = \sqrt{\dfrac{25}{4}}=\dfrac{5}{2}$ 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 : $d = \sqrt{(9-3)^2 + (\dfrac{15}{2}-3)^2} = \sqrt{6^2 + \dfrac{9}{2}^2} = \sqrt{36 + \dfrac{81}{4}} = \dfrac{15}{2}$ Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii : If $d > r_1 + r_2$, the circles are separate, and there are 4 common tangents.

If $d = r_1 + r_2$, the circles are externally tangent, and there are 3 common tangents.

If $0 < d < |r_1 - r_2|$, one circle is inside the other, and there are no common tangents.

If $d = |r_1 - r_2|$, the circles are internally tangent, and there is 1 common tangent.

If $d < |r_1 - r_2|$, one circle is completely inside the other, and there are no common tangents.

In this case : $d = \dfrac{15}{2}$ $r_1 = \dfrac{5}{2}$ $r_2 = 5$ Now, let's check the conditions : $r_1 + r_2 = \dfrac{5}{2} + 5$ = $\dfrac{15}{2}$ Since $d = \dfrac{15}{2} = r_1 + r_2$, the circles touch each other externally, and there are 3 common tangents.

First find point A by solving $4 x+3 y=1$ and $3 x-4 y=32$ After solving, point $\mathrm{A}$ is $(4,-5)$ Centre $(\alpha, \beta)$ lie on $4 x+3 y=1$

Now distance from centre to line $3 x-4 y-32=$ 0 and $3 x+4 y-24=0$ are equal.

After solving $\alpha=1$ and $\alpha=\dfrac{28}{3}$ For $\alpha=1$, centre $(1,-1) \Rightarrow$ radius $=5$ For $\alpha=\dfrac{28}{3}$, centre $\left(\dfrac{28}{3}, \dfrac{-109}{2}\right)$

Hence, $\alpha=1, \beta=-1, \mathrm{r}=5$

If two circles intersect at two distinct points

3

We have, equation of circle is $x^2+y^2=16$ Let any point on the circle $x^2+y^2=4^2$ is $B(4 \cos \theta, 4 \sin \theta)$ and $A(1,2)$ Let $\mathrm{P}$ be $(h, k)$ which divides $\mathrm{AB}$ in $3: 2$ So

$\Rightarrow \cos \theta=\dfrac{5 h-2}{12}$ and $\sin \theta=\dfrac{5 k-4}{12}$ As, $\cos ^2 \theta+\sin ^2 \theta=1,\left(\dfrac{5 h-2}{12}\right)^2+\left(\dfrac{5 k-4}{12}\right)^2=1$

$\therefore$ Locus of point $P$ is $\left(x-\dfrac{2}{5}\right)^2+\left(y-\dfrac{4}{5}\right)^2=\dfrac{12^2}{5^2}$ which is equation of circle with centre $\left(\dfrac{2}{5}, \dfrac{4}{5}\right)$ Hence, $A C=\sqrt{\left(1-\dfrac{2}{5}\right)^2+\left(2-\dfrac{4}{5}\right)^2}=\dfrac{\sqrt{45}}{5}=\dfrac{3 \sqrt{5}}{5}$

Given, length of $A B=\lambda$ So, $A C=\dfrac{\lambda}{2}$ and $A M=\dfrac{2 \lambda}{5}$

Now, from $\triangle O C M$, we get

So, radius is $\dfrac{\sqrt{19} \lambda}{5}$.

Centre $(3,-2)$ Equation of circumcircle is

Since $\left(\alpha, \dfrac{1}{2}\right)$ is on the circle