The point, the length of the tangents from which to the following three circles $x^{2} + y^{2} + 4 x + 6 y - 19 = 0$ , $x^{2} + y^{2} - 2 x - 2 y - 5 = 0$ and $x^{2} + y^{2} - 9 = 0$ are equal is

(2, - 1)

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(2, - 2)

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(1, 1)

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(1, - 2)

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Solution

The correct option is C $(1, 1)$
The required point will be the radical centre
$∴$ The common chord $R_{12}$ of the first two circle is
$S_{1} - S_{3} = 0 \Rightarrow 6 x + 8 y - 14 = 0 \Rightarrow 3 x + 4 y = 7$
The common chord $R_{23}$ of the second and third circles is
$S_{2} - S_{3} = 0 \Rightarrow - 2 x - 2 y + 4 = 0 \Rightarrow x + y = 2$
Solving these two, the point of intersection of $R_{12}$ and $R_{23}$ is $(1, 1)$

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The point, the length of the tangents from which to the following three circles x2+y2+4x+6y−19=0,x2+y2−2x−2y−5=0 and x2+y2−9=0 are equal is

The point, the length of the tangents from which to the following three circles $x^{2} + y^{2} + 4 x + 6 y - 19 = 0$ , $x^{2} + y^{2} - 2 x - 2 y - 5 = 0$ and $x^{2} + y^{2} - 9 = 0$ are equal is