If a chord is drwan through any point $P$ of the circle $x^{2} + y^{2} + 4 x - 6 y - 12 = 0$ become secant for circle $x^{2} + y^{2} + 4 x - 6 y + 4 = 0$ intersecting it at points at $A$ and $B$ , then $P A \cdot P B$ is

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Solution

Circle $C_{1} \equiv (- 2, 3), r_{1} = 5$
$C_{2} \equiv (- 2, 3), r_{1} = 3$
Hence circles are concentric circles

We know that $P A \cdot P B = P T^{2}$
In $△ P T C_{1}$
$P C_{1}^{2} = P T^{2} + T C_{1}^{2} \Rightarrow r_{1}^{2} = P T^{2} + r_{2}^{2} \Rightarrow P T^{2} = 16$

$∴ P A \cdot P B = 16$

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