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The point where the 'line $4 x - 3 y + 7 = 0$ touches the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0 is$

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$(- 1, - 1)$

$(1, - 1)$

$(- 1, 1)$

$(1, 1)$

answer is C.

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Detailed Solution

For the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ Centre C = (3, - 2)

Let p (h ,t) be the point of contact of 4x- 3y +7 = 0 ) $\to$ (l) with the circle

$∴ \frac{h - 3}{4} = \frac{k + 2}{- 3} = \frac{- (12 + 6 + 7)}{16 + 9} = - 1 \Rightarrow h - 3 = - 4, k + 2 = 3 \Rightarrow h = - 1, k = 1$

$∴$ Point of contact P = (- I, l )

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