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The inverse point of $(1,2)$ with respect to the circle $x^2+y^2-4 x-6 y+9=0$ is
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The correct answer is:
$(0,1)$
Inverse point of $P(1,2)$ w.r.t. the circle is the foot of the perpendicular of $P$ on the polar of $P$. Given circle is $x^2+y^2-4 x-6 y+9=0$
Polar of $P(1,2)$ is
$\begin{array}{rrrr} & & x \cdot 1+y \cdot 2-2(x+1)-3(y+2)+9=0 \\ \Rightarrow & x+2 y-2 x-2-3 y-6+9=0 \\ \Rightarrow & x+y-1=0\end{array}$
Verifying inverse point of $P$ is $(0,1)$.
Polar of $P(1,2)$ is
$\begin{array}{rrrr} & & x \cdot 1+y \cdot 2-2(x+1)-3(y+2)+9=0 \\ \Rightarrow & x+2 y-2 x-2-3 y-6+9=0 \\ \Rightarrow & x+y-1=0\end{array}$
Verifying inverse point of $P$ is $(0,1)$.
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