If $x = 7$ touches the circle $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ , then the coordinates of the point of contact are

(7, 3)

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(7, 4)

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(7, 8)

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(7, 2)

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Solution

The correct option is C $(7, 3)$
Given $x = 7$ touches circle $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$
Substitute $x = 7$ in the equation
$7^{2} + y^{2} - 4 (7) - 6 y - 12 = 0$
$49 + y^{2} - 28 - 6 y - 12 = 0$
$y^{2} - 6 y + 9 = 0$
$(y - 3)^{2} = 0$
$y = 3$
The point of contact is $(7, 3)$ .

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