The equations of the tangents to the circle x | vedclass

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Question

(A) The given circle is $x^2+y^2=36$,which has center $(0,0)$ and radius $r=6$. The given line is $5x+y=2$,which can be written as $y=-5x+2$. The slop

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$x-5y \pm 6\sqrt{26}=0$

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(A) The given circle is $x^2+y^2=36$,which has center $(0,0)$ and radius $r=6$. The given line is $5x+y=2$,which can be written as $y=-5x+2$. The slope of this line is $m_1=-5$. The tangent is perpendicular to this line,so its slope $m$ must satisfy $m 	imes (-5) = -1$,which gives $m = \frac{1}{5}$. The equation of a line with slope $m$ is $y = mx + c$,or $mx - y + c = 0$. Here,$\frac{1}{5}x - y + c = 0$,which simplifies to $x - 5y + 5c = 0$. The distance from the center $(0,0)$ to the tangent line $x - 5y + 5c = 0$ must be equal to the radius $r=6$. Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$,we have $6 = \frac{|0 - 0 + 5c|}{\sqrt{1^2+(-5)^2}}$. $6 = \frac{|5c|}{\sqrt{26}} \implies |5c| = 6\sqrt{26} \implies 5c = \pm 6\sqrt{26}$. Substituting $5c$ back into the equation $x - 5y + 5c = 0$,we get $x - 5y \pm 6\sqrt{26} = 0$.

The equations of the tangents to the circle $x^2+y^2=36$ which are perpendicular to the line $5x+y=2$ are:

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