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If O is the origin and OP and OQ are the tangents from the origin to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ , then the circumcenter of triangle OPQ is

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$(3, - 2)$

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

$(\frac{3}{4}, - \frac{1}{2})$

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

answer is B.

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Detailed Solution

Clearly, OPCQ is a cyclic quadrilateral. Then the circumcircle of $∆$ OPQ passes through the point C.
For this circle, OC is diameter. Then the center is the midpoint of OC, which is ( $\frac{3}{2}$ , -1).

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