The equations of the tangents to the circle x | vedclass

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(A) The equation of the circle is $x^2+y^2=36$,so the radius $r=6$ and the center is $(0,0)$.Any line perpendicular to $5x+y-2=0$ will have the form $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2=36$,so the radius $r=6$ and the center is $(0,0)$.Any line perpendicular to $5x+y-2=0$ will have the form $x-5y+k=0$.The distance from the center $(0,0)$ to the tangent line $x-5y+k=0$ 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 get $6 = \frac{|1(0)-5(0)+k|}{\sqrt{1^2+(-5)^2}}$.$6 = \frac{|k|}{\sqrt{26}}$,which implies $|k| = 6\sqrt{26}$.Thus,$k = \pm 6\sqrt{26}$.The equations of the tangents are $x-5y \pm 6\sqrt{26}=0$.

The equations of the tangents to the circle $x^2+y^2=36$ which are perpendicular to the line $5x+y-2=0$ are

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