Let P and Q be the inverse points with respec | vedclass

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(D) The equation of the circle is $x^2+y^2-4x-6y+k=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-3$. The centre $C$ is $(-g, -f) = (2, 3)$

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(D) The equation of the circle is $x^2+y^2-4x-6y+k=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-3$. The centre $C$ is $(-g, -f) = (2, 3)$.Given $CP \cdot CQ = r^2 = 4$,so $r^2 = 4$.For inverse points $P$ and $Q$ with respect to a circle with centre $C(h, k)$ and radius $r$,the points $C, P, Q$ are collinear and $CP \cdot CQ = r^2$.The vector relation is $\vec{CQ} = \frac{r^2}{CP^2} \vec{CP}$.Here,$P = (1, 2)$ and $C = (2, 3)$.$CP^2 = (1-2)^2 + (2-3)^2 = (-1)^2 + (-1)^2 = 1 + 1 = 2$.Since $CP \cdot CQ = 4$,we have $CQ = \frac{4}{CP} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The vector $\vec{CP} = P - C = (1-2, 2-3) = (-1, -1)$.Then $\vec{CQ} = \frac{r^2}{CP^2} \vec{CP} = \frac{4}{2} (-1, -1) = 2(-1, -1) = (-2, -2)$.Thus,$Q - C = (-2, -2) \Rightarrow Q = C + (-2, -2) = (2-2, 3-2) = (0, 1)$.So,$a=0$ and $b=1$.Therefore,$2a = 2(0) = 0$.

Let $P$ and $Q$ be the inverse points with respect to the circle $S \equiv x^2+y^2-4x-6y+k=0$ and $C$ be the centre of the circle $S=0$ such that $CP \cdot CQ=4$. If $P=(1,2)$ and $Q=(a, b)$,then $2a=$

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