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

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(D) The given equation of the circle is $x^2+y^2+2x-12y-132=0$. Completing the square,we get $(x+1)^2+(y-6)^2 = 132+1+36 = 169 = 13^2$. Thus,the cente

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$5x-12y+246=0$

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(D) The given equation of the circle is $x^2+y^2+2x-12y-132=0$. Completing the square,we get $(x+1)^2+(y-6)^2 = 132+1+36 = 169 = 13^2$. Thus,the center is $(-1, 6)$ and the radius $r = 13$. The slope of the line $12x+5y+k=0$ is $m_1 = -\frac{12}{5}$. The slope of the tangent perpendicular to this line is $m = \frac{5}{12}$. The equation of the tangent can be written as $5x-12y+c = 0$. The perpendicular distance from the center $(-1, 6)$ to the tangent is equal to the radius $13$: $\frac{|5(-1)-12(6)+c|}{\sqrt{5^2+(-12)^2}} = 13$ $\frac{|-5-72+c|}{13} = 13$ $|c-77| = 169$ $c-77 = 169 \Rightarrow c = 246$ or $c-77 = -169 \Rightarrow c = -92$. Thus,the equations of the tangents are $5x-12y+246=0$ and $5x-12y-92=0$.

The equation of a tangent to the circle $x^2+y^2+2x-12y-132=0$ which is perpendicular to the line $12x+5y+k=0$ is

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