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Let $P$ and $Q$ be the inverse points with respect to the circle $S \equiv x^2+y^2-4 x-6 y+k=0$ and $C$ be the centre of the circle $S=0$ such that $C P$. $C Q=4$. If $P=(1,2)$ and $\mathrm{Q}=(\mathrm{a}, \mathrm{b})$, then $2 \mathrm{a}=$
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We have $C P \cdot C Q=r^2=4 \Rightarrow r=2$
Centre of given circle $x^2+y^2-4 x-6 y+k=0$ $=C(2,3)$.
We know that inverse of point $P(\alpha, \beta)$ with respect to circle with centre $(h, k)$ and radius $r$ is the point $(a, b)$ then $a=\lambda(\alpha-h)+h$ and $b=\lambda(\beta-k)+k$
$\begin{aligned}
& \text { where } \lambda=\frac{r^2}{(\alpha-h)^2+(\beta-k)^2} \\
& \therefore \quad \lambda=\frac{4}{(1-2)^2+(2-3)^2}=2 \\
& \therefore \quad a=2(1-2)+2=-2+2=0 .
\end{aligned}$
Centre of given circle $x^2+y^2-4 x-6 y+k=0$ $=C(2,3)$.
We know that inverse of point $P(\alpha, \beta)$ with respect to circle with centre $(h, k)$ and radius $r$ is the point $(a, b)$ then $a=\lambda(\alpha-h)+h$ and $b=\lambda(\beta-k)+k$
$\begin{aligned}
& \text { where } \lambda=\frac{r^2}{(\alpha-h)^2+(\beta-k)^2} \\
& \therefore \quad \lambda=\frac{4}{(1-2)^2+(2-3)^2}=2 \\
& \therefore \quad a=2(1-2)+2=-2+2=0 .
\end{aligned}$
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