The equations of the tangents to the circle x | vedclass

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(C) The given 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

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

Vedclass · Answer

(C) The given 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 $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 2^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$. Any line parallel to $4x + 3y + 5 = 0$ is of the form $4x + 3y + k = 0$. Since this line is a tangent to the circle,the perpendicular distance from the center $(3, -2)$ to the line must be equal to the radius $r = 5$. $\frac{|4(3) + 3(-2) + k|}{\sqrt{4^2 + 3^2}} = 5$ $\frac{|12 - 6 + k|}{5} = 5$ $|6 + k| = 25$ $6 + k = 25$ or $6 + k = -25$ $k = 19$ or $k = -31$. Thus,the equations of the tangents are $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

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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