lf a circle having the point $(- 1, 1)$ as centre and touches the straight line $x + 2 y + 9 = 0$ , then the coordinate of the point (s) of contact are

(- 3, 3)

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(- 3, - 3)

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(0, 0)

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(\frac{7}{3} \frac{- 7}{3})

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Solution

The correct option is B $(- 3, - 3)$
Given that $x + 2 y + 9 = 0$ is a tangent to a circle having center at (-1, 1) .
Let point of contact is P (h, k). Then foot of perpendicular from (-1, 1) to $x + 2 y + 9 = 0$ is P (h, k).
$\frac{h + 1}{1} = \frac{k - 1}{2} = - (\frac{10}{1^{2} + 2^{2}})$
$h = - 3$ and $k = - 3$
then $P (- 3, - 3)$ is a point of contact.

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lf a circle having the point (−1,1) as centre and touches the straight line x+2y+9=0, then the coordinate of the point (s) of contact are

The correct option is B (−3,−3)Given that x+2y+9=0 is a tangent to a circle having center at (-1, 1) .Let point of contact is P (h, k). Then foot of perpendicular from (-1, 1) to x+2y+9=0 is P (h, k).h+11=k−12=−(1012+22)h=−3 and k=−3then P(−3,−3) is a point of contact.

lf a circle having the point $(- 1, 1)$ as centre and touches the straight line $x + 2 y + 9 = 0$ , then the coordinate of the point (s) of contact are