From a point P(-4, 0),two tangents are drawn | vedclass

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(D) The given circle is $S: x^2 + y^2 - 4x - 6y - 12 = 0$. The center is $C(2, 3)$ and radius $r = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = 5$.

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

Vedclass · Answer

(D) The given circle is $S: x^2 + y^2 - 4x - 6y - 12 = 0$. The center is $C(2, 3)$ and radius $r = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = 5$. Let $P = (-4, 0)$. The circle passing through $P, A$,and $B$ has the segment $PC$ as its diameter,where $C$ is the center of the original circle and $P$ is the external point. The midpoint of $PC$ is the center of the new circle: $M = \left(\frac{-4 + 2}{2}, \frac{0 + 3}{2}\right) = \left(-1, \frac{3}{2}\right)$. The equation of the circle with diameter $PC$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$,which is $(x + 4)(x - 2) + (y - 0)(y - 3) = 0$. Expanding this,we get $x^2 + 2x - 8 + y^2 - 3y = 0$,or $x^2 + y^2 + 2x - 3y - 8 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = 2 \implies g = 1$ and $2f = -3 \implies f = -\frac{3}{2}$. Thus,$(g, f) = \left(1, -\frac{3}{2}\right)$.

From a point $P(-4, 0)$,two tangents are drawn to the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ touching the circle at $A$ and $B$. If the equation of the circle passing through $P, A$,and $B$ is $x^2 + y^2 + 2gx + 2fy + c = 0$,then $(g, f) =$

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