Circle

Given, from point $P(1,1)$, tangent $P A$ and $P B$ is drawn

$ \begin{aligned} & \therefore \quad P A=\sqrt{S_1} \\ & \Rightarrow \quad P A=\sqrt{1+1+g+g-2}=\sqrt{2 g} \\ & \text { Radius of circle }=\sqrt{\frac{g^2}{4}+\frac{g^2}{4}+2} \\ & \Rightarrow \quad A C=\frac{1}{\sqrt{2}} \sqrt{g^2+4} \end{aligned} $

$ \begin{aligned} & \therefore \text { Area of } \triangle P A C =\frac{1}{2} P A \times A C \\ & =\frac{1}{2} \times \sqrt{2 g} \times \frac{1}{\sqrt{2}} \sqrt{g^2+4} \\ & \therefore \text { Area of } P A C B =2 \times \text { area of } \triangle P A C \\ & =\frac{1}{\sqrt{2}} \sqrt{2 g} \sqrt{g^2+4} \\ & =\sqrt{g^3+4 g} \end{aligned} $

Given circle,

$ S_1: x^2+y^2-8 x-6 y+21=0 $

and $S_2: x^2+y^2-2 y-15=0$

Circle $S_1$ : Centre $C_1(4,3)$, radius

$ r_1=\sqrt{16+9-21}=2 $

$S_2$ : Centre $C_2(0,1)$, radius

$ r_2=\sqrt{1+15}=4 $

Now $C_1 C_2=\sqrt{4^2+2^2}=\sqrt{20}$

$ \therefore C_1 C_2

⇒ circle $C_1$ and $C_2$ put with other.

Let point of intersection of tangent on the circle $S_1$ and $S_2$ is $P$.

Now, point $P$ divide line joining $C_2 C_1$ in the ratio of their radii externally

$ \therefore \quad \frac{C_2 P}{C_1 P}=\frac{r_2}{r_1} \Rightarrow \frac{C_2 P}{C_1 P}=\frac{4}{2}=\frac{2}{1} $

∴ By using external division formula

$ \begin{aligned} & =\left(\frac{m_1 n_2-m_2 n_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}\right) \\ & =\left(\frac{2 \times 4-1 \times 0}{2-1}, \frac{2 \times 3-1 \times 1}{2-1}\right) \\ & =(8,5) \end{aligned} $

As radical axis of the two circles bisects the all common tangents of the circles.

Now, the mid-point of $A(1,1)$ and $B(0,1)$ is $M\left(\frac{1}{2}, 1\right)$ and mid-point of $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$ is $N\left(\frac{1}{5}, \frac{11}{5}\right)$

∴ Equation of required radical axis $M N$ is

$ \begin{aligned} y-1 & =\frac{\frac{11}{10}-1}{\frac{1}{5}-\frac{1}{2}}\left(x-\frac{1}{2}\right) \\ \Rightarrow \quad 3(y-1) & =-x+\frac{1}{2} \\ \Rightarrow \quad 2 x+6 y & =7 \end{aligned} $

Let the equation of the required circle be

$ x^2+y^2+2 g x+2 f y+c=0 $

and centre is $(-g,-f)$

Given circles,

$ C_1: x^2+y^2-2 x-4 y-4=0 $

Here, $g_1=-1, f_1=-2, c_1=-4$

$ C_2: x^2+y^2-10 x+12 y+52=0 $

Here, $g_2=-5, f_2=6, c_2=52$

Condition of two circles cuts orthogonally

$ 2 g_1 g_2+2 f_1 f_2=c_1+c_2 $

For $C_1: 2(g)(-1)+2(f)(-2)=C-4$

or $2 g+4 f=-c+4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For $C_2: 2(g)(-5)+2(f)(6)=c+52$

$ \begin{equation*} 10 g-12 f=-c-52\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{equation*} $

Subtracting on Eqs. (i) and (ii), we get

$ -8 g+16 f=56 $

or     $ g-2 f=-7 $

or     $ \begin{equation*} g=2 f-7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{equation*} $

Multiplying Eq. (i) by 5 and then subtracting

Eq. (ii), we get

or $ 8 f=-c+18 \text { or } c=-8 f+18 $

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

$ \begin{aligned} & =\sqrt{(2 f-7)^2+f^2(-8 f+18)} \\ & =\sqrt{5 f^2-20 f+31} \\ & =\sqrt{5\left(f^2-4 f\right)+31}=\sqrt{5(f-2)^2+11} \end{aligned} $

For minimum, value $f-2=0$

$ \Rightarrow \quad f=2 $

Putting the value of ' $f$ ' in Eq. (iii), we get

$ g=2 \times 2-7 \Rightarrow g=-3 $

Therefore, coordinate of centre is $(+3,-2)$.

A circle is passing through $(3,4),(3$, $2)$ and ( 1,4 ).

Clearly, $\triangle A B C$ is a right angle triangle - centre of circle is mid-point of $A C$ i.e., $O(2,3)$

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

Equation of circle in parameter form is $x=2+\sqrt{2} \cos \theta \Rightarrow y=3+\sqrt{2} \sin \theta$

$ \quad \begin{aligned} \therefore\,\,\,\,\,\, a & =2, b=3, r=\sqrt{2} \\ b^a \cdot r^a & =(3)^2(\sqrt{2})^2=9 \times 2=18 \end{aligned} $

Given equation of circle,

$ \begin{gathered} x^2+y^2-4 x-6 y+9=0 \\ R=P C=\sqrt{4+9-9}=2 \\ x^2+y^2-4 x-6 y+12=0 \\ r=Q C=\sqrt{4+9-12}=1 \end{gathered} $

In $\triangle P Q C$,

$ \cos \theta=\frac{Q C}{P C}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} $

∴ Area of $\triangle Q R C$,

$ \begin{aligned} & =\frac{1}{2}(C Q)^2 \times \sin 2 \theta=\frac{1}{2} \times 1 \times \sin \frac{2 \pi}{3} \\ & =\frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \end{aligned} $

Equation of tangent drawn from origin, ' $O$ ' of the circle is

$ \begin{aligned} & \quad x^2+y^2+2 g x+2 f y+g^2=0 \\ & \text { } \\ &or \,\,\,\,\,\,\,\,\,\,\,\,\, \quad T^2=S S_1 \\ & \therefore\left(0+0+2 g\left(\frac{x+0}{2}\right)+2 f \frac{(y+0)}{2}+g^2\right)^2 \\ & \quad=\left(x^2+y^2+2 g x+2 f y+g^2\right)\left(g^2\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow\left(g x+f y+g^2\right)^2 \\ & \quad=g^2\left(x^2+y^2\right)+2 g^3 x+2 f g^2 y+g^4 \\ & \begin{aligned} \Rightarrow & g^2 x^2+f^2 y^2+g^4+2 g^3 x +2 g^2 f y+2 g f x y \\ & =g^2 x^2+g^2 y^2+2 g^3 x+2 f g^2 y+g^4 \\ \Rightarrow & f^2 y^2+2 g^2 f y+2 g f x y=g^2 y^2+2 f g^2 y \\ \Rightarrow & y^2\left(g^2-f^2\right)-2 g f y x=0 \\ \Rightarrow & y\left[\left(g^2-f^2\right) y-2 g f x\right]=0 \\ \Rightarrow & y=0 \\ & \left(g^2-f^2\right) y-2 g f x=0 \end{aligned} \end{aligned} $

$2 x+y=0$ is equation of chord of circle,

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

$2 x+y=0$ is also diameter of the required circle

∵ Centre of circle $\left(\lambda-1, \frac{\lambda-6}{2}\right)$ lie on chord

$ \begin{aligned} \therefore & & 2(\lambda-1)+\frac{\lambda-6}{2} & =0 \\ \Rightarrow & & \lambda & =2 \end{aligned} $

∵ Equation of circle

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

$(-2,1)$ is lie on circle

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

Radical axis of circle

$ x^2+y^2+2 \alpha x+2 \beta y+c=0 $

and $\quad x^2+y^2+\frac{3}{2} x+4 y+c=0$ is

$ \begin{aligned} & \left(2 \alpha-\frac{3}{2}\right) x+(2 \beta-4) y=0 \\ & (4 \alpha-3) x+4(\beta-2) y=0 \end{aligned} $

$(4 \alpha-3) x+(\beta-8) y=0$ touches the circle

$ \therefore \quad 1=\left|\frac{(4 \alpha-3)(-1)+(3-8)(-1)}{\sqrt{(4 \alpha-3)^2+(4 \beta-8)^2}}\right| $

$ \begin{aligned} & \Rightarrow(4 \alpha-3)^2+(4 b-8)^2 \\ & \quad=(4 \alpha+4 \beta-11)^2 \\ & \Rightarrow 16 \alpha^2-24 \alpha+9+16 \beta^2-64 \beta+64 \\ & \quad=16 \alpha^2+16 \beta^2+121+32 \alpha \beta-88 \alpha-88 \beta \end{aligned} $

$ \begin{aligned} \Rightarrow 32 \alpha \beta-88 \alpha+24 \alpha- & 88 \beta+64 \beta \\ & =9+64-121 \\ 32 \alpha \beta-64 \alpha-24 \beta & =-48 \\ \Rightarrow 4 \alpha \beta-8 \alpha-3 \beta & =-6 \\ \Rightarrow 4 \alpha \beta-8 \alpha-3 \beta+10 & =-6+10 \\ & =4 \end{aligned} $

Given circle

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

Centre $(2,1)$

Equation of diameter of circle passes through origin.

$ x-2 y=0 $

Equation of circle through the end point of diameter of circle and line $x-2 y=0$ is

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

Since, this is passes through $(1,2)$

$ \quad \begin{aligned} \because \quad 1+4-4-4-4+\lambda(-3) & =0 \\ \lambda & =\frac{-7}{3} \end{aligned} $

Required equation of circle is

$ \begin{array}{r} x^2+y^2-4 x-2 y-4-\frac{7}{3}(x-2 y)=0 \\ 3\left(x^2+y^2\right)-19 x+8 y-12=0 \\ 3 x^2+3 y^2-19 x+8 y-12=0 \end{array} $

We have, ( $2, \alpha$ ) is mid-point of chord of circle

$ x^2+y^2-4 x+8 y+6=0 $

$\because(2, \alpha)$ inside the circle

$ \begin{aligned} & \therefore \quad 4+\alpha^2-8+8 \alpha+6 \leq 0 \\ & \alpha^2+8 \alpha+2 \leq 0 \\ & \alpha=\frac{-8 \pm \sqrt{64-8}}{2} \\ & \alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14} \\ & \alpha \in(-4-\sqrt{14},-4+\sqrt{14}) \end{aligned} $

$C_1$ and $C_2$ are external and internal similitude of the circle

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

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

Centre of circle

$ \begin{aligned} x^2+y^2-2 x+4 y+1 & =0 \text { is } \\ (1,-2), r_1 & =2 \end{aligned} $

Centre of circle

$ \begin{array}{r} x^2+y^2+4 x-6 y+4=0 \\ (-2,3), r_2=1 \end{array} $

Coordinate of $C_1=\frac{-5}{3}$

$ \left(\frac{2(-2)-1}{2-1}, \frac{6+2}{2-1}\right) \equiv(-5,8) $

Coordinate of

$ \begin{aligned} C_2= & \left(\frac{2(-2)+1}{2+1}, \frac{6-2}{2+1}\right) \equiv\left(-1, \frac{4}{3}\right) \\ & C_1 C_2=\sqrt{(-5+1)^2+\left(8-\frac{4}{3}\right)^2} \\ & C_1 C_2=\sqrt{16+\frac{400}{9}} \\ \Rightarrow & C_1 C_2=\sqrt{\frac{544}{9}}=\frac{4 \sqrt{34}}{3} \end{aligned} $

$ \begin{aligned} &\text { Here, } C_1 C_2=2 r\\ &\begin{aligned} \therefore \quad \frac{4}{3} \sqrt{34} & =2 r \\ \frac{9}{2} r & =3 \sqrt{34} \end{aligned} \end{aligned} $

$x^2+y^2+2 g x+2 f y+c=0$ cuts

orthogonally the circle

$ \begin{aligned} & x^2+y^2-4 x-6 y+11=0 \text { and } \\ & \quad x^2+y^2-10 x-4 y+21=0 \end{aligned} $

$ \begin{array}{lll} \therefore & & -4 g-6 f=c+11\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \text { and } & & -10 g-4 f=c+21 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \text { } \end{array} $

Centre of circle $x^2+y^2+2 g x+2 f y+c=0$ is lie in bisector of angle between the positive coordinate axis

$ \therefore \quad g=f \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eqs. (i) and (iii), we get

$ \begin{array}{r} -4 f-6 f=c+11 \\ -10 f=c+11 ...(iv)\end{array} $

From Eqs. (ii) and (iii), we get

$ -14 f=c+21 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

From Eqs. (iv) and (v), we get

$ \begin{array}{rlrl} & g=f=\frac{-5}{2}, c=14 \\ \therefore & 2 g+2 f+c-5-5+14= & 4 \end{array} $

Given $S_1=x^2+y^2=16$

Centre $(0,0) r=4$

Clearly, the diameter of $S_1$ will be common chord. Let $P Q$ be common chord and centre of $S_2$ be $(h, k)$

We have $A P=5, P B=4$,

$ \therefore \quad A B=3 $

Where $B=(0,0)$

Slope of $P Q=3 / 4$

∴ Slope of $A B=\frac{-4}{3}$

By parameter equation

$ \begin{aligned} \frac{h-0}{\frac{-3}{5}} & =\frac{k-0}{\frac{4}{5}}= \pm 3 \\ x=\mp \frac{9}{5}, y & =\frac{ \pm 12}{5} \end{aligned} $

∴ Centre $\left(\frac{-9}{5}, \frac{12}{5}\right)$ and $\left(\frac{9}{5}, \frac{-12}{5}\right)$

Equation of circle passing through the points $(1,1),(2,-1)$ and $(3,2)$ is

$ \begin{aligned} & \left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ 1+1 & 1 & 1 & 1 \\ 4+1 & 2 & -1 & 1 \\ 9+4 & 3 & 2 & 1 \end{array}\right|=0 \\ & \left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ 2 & 1 & 1 & 1 \\ 5 & 2 & -1 & 1 \\ 13 & 3 & 2 & 1 \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \left|\begin{array}{cccc} x^2+y^2-2 & x-1 & y-1 & 0 \\ -3 & -1 & 2 & 0 \\ -8 & -1 & -3 & 0 \\ 13 & 3 & 2 & 1 \end{array}\right|=0 \\ & \left|\begin{array}{ccc} x^2+y^2-2 & x-1 & y-1 \\ -3 & -1 & 2 \\ -8 & -1 & -3 \end{array}\right|=0 \\ & \left(x^2+y^2-2\right)(3+2)-(x-1)(9+16)+(y-1)(3-8)=0 \\ & \quad \begin{array}{l} 5\left(x^2+y^2-2\right)-25(x-1)-5(y-1)=0 \\ x^2+y^2-2-5(x-1)-(y-1)=0 \\ x^2+y^2-2-5 x+5-y+1=0 \\ x^2+y^2-5 x-y+4=0 \end{array} \end{aligned} $

Now, by putting all the points, we see $\left(\frac{5}{2}+\sqrt{\frac{5}{2}}, \frac{1}{2}\right),\left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ passes through above circle.

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

Now, equation of circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is

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

Now, polar of $P$ with respect to the circle $x^2+y^2-2 x r-2 y r-r^2=0$ is

$ \begin{aligned} & x x_1+y y_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0 \\ & \Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r \end{aligned} $

But it is given that polar of $P$ with respect to above circle is $x+2 y=4 r$

$ \begin{aligned} & \therefore \quad \frac{x_1-r}{1}=\frac{y_1-r}{2}=\frac{x_1+y_1-r}{4} \\ & \Rightarrow \quad x_1-r=\frac{x_1+y_1-r}{4} \\ & \text { and } x_1-r=\frac{y_1-r}{2} \\ & \Rightarrow \quad 4 x_1-4 r=x_1+y_1-r \text { and } \\ & 2 x_1-2 r=y_1-r \\ & \Rightarrow \quad 3 x_1-3 r=y_1 \text { and } 2 x_1-r=y_1 \\ & \therefore \quad 3 x_1-3 r=2 x_1-r \\ & \Rightarrow \quad x_1=2 r \text { and } y_1=2 x_1-r \\ & \quad 4 r-r=3 r \\ & \quad \quad P \text { is }(2 r, 3 r) \end{aligned} $

The centres of the given circles are $C_1(1,3+\sqrt{7})$ and $C_2(4,3)$ and corresponding radii are

$ \begin{aligned} r_1 & =\sqrt{1^2+(3+\sqrt{7})^2-(8+6 \sqrt{7})}=3 \\ \text { and } r_2 & =\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2} \end{aligned} $

Now, $C_1 C_2=\sqrt{(4-1)^2+(3-3-\sqrt{7})^2}=4$

Clearly, $C_1 C_2

$ \begin{aligned} & \Rightarrow \quad 4<3+\sqrt{25-k^2} \Rightarrow 1<\sqrt{25-k^2} \\ & \Rightarrow \quad 1<25-k^2 \Rightarrow k^2<24 \end{aligned} $

$ \therefore k=0, \pm 1, \pm 2, \pm 3, \pm 4 \quad[\because k \in \mathbf{Z}] $

Let the equation of circle $S$ is

$ \begin{aligned} & S=x^2+y^2+2 g x+2 f y+c=0 \\ & C_1=x^2+y^2-8 x-2 y+16=0 \\ & C_2=x^2+y^2-4 x-4 y-1=0 \end{aligned} $

$S$ is orthognoally to $C_1$ and $C_2$

$ \begin{array}{lrl} \therefore & -8 g-2 f & =C+16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & -4 g-4 f & =C-1 \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array} $

From Eqs. (i) and (ii),

$ -4 g+2 f=17 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Common chord of $S$ and $C_1=$

$ (2 g+8) x+(2 f+2) y+c-16=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Given common chord is

$ 2 x+13 y-15=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

Eqs. (iv) and (v) are coinside

$ \begin{array}{ll} \therefore & \frac{2 g+8}{2}=\frac{2 f+2}{13}=\frac{c-16}{-15} \\ \Rightarrow & 13 g-2 f=-50 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} $

Solving Eqs. (iii) and (vi), we get

$ g=\frac{11}{3} f=\frac{-7}{6} $

∴ Centre of circle, $S=\left(\frac{-11}{3}, \frac{7}{6}\right)$

We have,

$ \begin{aligned} & S_1: x^2+y^2+k x-4 y-1=0 \\ & S_2: 3 x^2+3 y^2-14 x+23 y-15=0 \\ & \quad=x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5=0 \end{aligned} $

Since, $S_1$ and $S_2$ are orthogonal

$ \begin{array}{ll} \therefore & 2\left[\left(\frac{-k}{2}\right)\left(\frac{7}{3}\right)+(2)\left(\frac{-23}{6}\right)\right]=-1-5 \\ \Rightarrow & 2\left[\frac{-7 k}{6}-\frac{46}{6}\right]=-6 \\ \Rightarrow & \frac{-7 k}{6}-\frac{46}{6}=-3 \\ \Rightarrow & \frac{7 k}{6}=3-\frac{46}{6} \\ \Rightarrow & \frac{7 k}{6}=\frac{18-46}{6} \\ \Rightarrow & 7 k=-28 \\ \Rightarrow & k=-4 \\ \therefore & S_1: x^2+y^2-4 x-4 y-1=0 \\ & S_2: x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5=0 \end{array} $

Equation of circle passing through intersection of $S_1$ and $S_2$ is

$ \begin{aligned} & x^2+y^2-4 x-4 y-1 \\ & +\lambda\left(x^2+y^2-\frac{14 x}{3}+\frac{23}{3} y-5\right)=0 \end{aligned} $

Since above circle passes through $(-1,-1)$, so

$ 1+1+4+4-1+\lambda\left(1+1+\frac{14}{3}-\frac{23}{3}-5\right)=0 $

$ \begin{aligned} &9-6 \lambda=0 \Rightarrow \lambda=\frac{3}{2}\\ &\text { ∴ Equation of required circle is }\\ &\begin{aligned} & x^2+y^2-4 x-4 y-1+\frac{3}{2} \\ & \left(x^2+y^2-\frac{14 x}{3}+\frac{23 y}{3}-5\right)=0 \\ & \Rightarrow \quad 2 x^2+2 y^2-8 x-8 y-2+3 x^2+3 y^2 -14 x+23 y-15=0 \\ & \Rightarrow \quad 5 x^2+5 y^2-22 x+15 y-17=0 \end{aligned} \end{aligned} $