The line 4x + 4y – 11 = 0 intersects the circle $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ at A and B. The point of intersection of the tangents at A, B is

(–1, – 2)

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(1, 2)

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(–1, 2)

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(1, –2)

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Solution

The correct option is A (–1, – 2)
The point of intersection of the tangents at A, B is the pole of 4x + 4y – 11 = 0 with respect to $x^{2} + y^{2} - 6 x - 4 y + 4 = 0 \equiv (x - 3)^{2} + (y - 2)^{2} = 3^{2}$
$∴ R e q u i r e d p o i n t = (3 - \frac{9 (4)}{12 + 8 - 11}, 2 - \frac{9 (4)}{12 + 8 - 11}) = (- 1, - 2)$

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