The equations of the tangents to the circle x | vedclass

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(C) The equation of the circle is $x^{2}+y^{2}-6x+4y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=2, c=-12$. The center is $(-g, -f

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$4x+3y+19=0, 4x+3y-31=0$

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(C) The equation of the circle is $x^{2}+y^{2}-6x+4y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=2, c=-12$. The center is $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{9+4+12} = \sqrt{25} = 5$. Any line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$. The distance from the center $(3, -2)$ to the tangent line is equal to the radius: $\frac{|4(3)+3(-2)+k|}{\sqrt{4^{2}+3^{2}}} = 5$ $\frac{|12-6+k|}{5} = 5$ $|6+k| = 25$ $6+k = 25$ or $6+k = -25$ $k = 19$ or $k = -31$. Thus,the equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

The equations of the tangents to the circle $x^{2}+y^{2}-6x+4y=12$,which are parallel to the straight line $4x+3y+5=0$,are

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