The equation of the tangent to the circle x^2 | vedclass

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Question

(A) The equation of the circle is $x^2 + y^2 - 22x - 4y + 25 = 0$. The center is $(11, 2)$ and the radius $r = \sqrt{11^2 + 2^2 - 25} = \sqrt{121 + 4

Vedclass · Accepted Answer

$12x - 5y + 8 = 0, 12x - 5y = 252$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 - 22x - 4y + 25 = 0$. The center is $(11, 2)$ and the radius $r = \sqrt{11^2 + 2^2 - 25} = \sqrt{121 + 4 - 25} = \sqrt{100} = 10$.The line perpendicular to $5x + 12y + 8 = 0$ is of the form $12x - 5y + k = 0$.Since the line is a tangent,the perpendicular distance from the center $(11, 2)$ to the line $12x - 5y + k = 0$ must be equal to the radius $r = 10$.$\frac{|12(11) - 5(2) + k|}{\sqrt{12^2 + (-5)^2}} = 10$$\frac{|132 - 10 + k|}{\sqrt{144 + 25}} = 10$$\frac{|122 + k|}{13} = 10$$|122 + k| = 130$$122 + k = 130 \implies k = 8$$122 + k = -130 \implies k = -252$Thus,the equations of the tangents are $12x - 5y + 8 = 0$ and $12x - 5y - 252 = 0$,which is $12x - 5y = 252$.

The equation of the tangent to the circle $x^2 + y^2 - 22x - 4y + 25 = 0$ which is perpendicular to the line $5x + 12y + 8 = 0$ is:

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