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The normal to a circle $S=0$ at $P(1,3)$ is $x+2 y=7$ and it has another normal at $Q(3,5)$ which is the polar of the point $A\left(7,-\frac{1}{2}\right)$ with respect to the circle $x^2+y^2-4 x+6 y-12=0$. Then, the equation of the circle $S=0$ is
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Verified Answer
The correct answer is:
$x^2+y^2-10 x-2 y+6=0$
Polar of the point $A\left(7, \frac{-1}{2}\right)$ with respect to the circle $x^2+y^2-4 x+6 y-12=0$
$$
\begin{array}{lrl}
\Rightarrow & 7 x-\frac{1}{2} y-2(x+7)+3\left(y-\frac{1}{2}\right)-12=0 \\
\Rightarrow & 14 x-y-4 x-28+6 y-3-24=0 \\
\Rightarrow & 10 x+5 y-55=0 \\
\Rightarrow & 2 x+y-11=0
\end{array}
$$
Normal of circle $s=0$ at $(1,3)$ is $x+2 y=7$
at $(3,5)$ is $2 x+y=11$
Solving Eqs. (i) and (ii), we get
$$
x=5, y=1
$$
$\therefore$ Centre of circle $s=0$ is $(5,1)$.
$\therefore$ Equation of circle is
$$
\begin{aligned}
& (x-5)^2+(y-1)^2=\left\{\sqrt{(5-1)^2+(1-3)^2}\right\}^2 \\
& \Rightarrow x^2-10 x+25+y^2-2 y+1=20 \\
& \Rightarrow \quad x^2+y^2-10 x-2 y+6=0
\end{aligned}
$$
$$
\begin{array}{lrl}
\Rightarrow & 7 x-\frac{1}{2} y-2(x+7)+3\left(y-\frac{1}{2}\right)-12=0 \\
\Rightarrow & 14 x-y-4 x-28+6 y-3-24=0 \\
\Rightarrow & 10 x+5 y-55=0 \\
\Rightarrow & 2 x+y-11=0
\end{array}
$$
Normal of circle $s=0$ at $(1,3)$ is $x+2 y=7$
at $(3,5)$ is $2 x+y=11$
Solving Eqs. (i) and (ii), we get
$$
x=5, y=1
$$
$\therefore$ Centre of circle $s=0$ is $(5,1)$.
$\therefore$ Equation of circle is
$$
\begin{aligned}
& (x-5)^2+(y-1)^2=\left\{\sqrt{(5-1)^2+(1-3)^2}\right\}^2 \\
& \Rightarrow x^2-10 x+25+y^2-2 y+1=20 \\
& \Rightarrow \quad x^2+y^2-10 x-2 y+6=0
\end{aligned}
$$
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