The line $y = x + 1$ is a tangent to the circle $x^{2} + y^{2} + 2 x + 2 y + c = 0$ . The corresponding point of contact is

(\frac{- 3}{2}, \frac{- 1}{2})

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(- 1, 0)

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(2, 3)

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(1, 2)

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Solution

The correct option is B $(\frac{- 3}{2}, \frac{- 1}{2})$
The center of the circle is $C (- 1, - 1)$ .
The foot of perpendicular from center C to the tangent is the point of contact $P (h, k)$ .
$\Rightarrow \frac{h + 1}{1} = \frac{k + 1}{- 1} = - (\frac{- 1 + 1 + 1}{1 + 1})$
$\Rightarrow h = \frac{- 3}{2}, k = \frac{- 1}{2}$

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