The equations of tangents to the circle x^2 + | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 22x - 4y + 25 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -11, f = -2, c = 25$. Th

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

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

The equations of tangents to the circle $x^2 + y^2 - 22x - 4y + 25 = 0$ which are perpendicular to the line $5x + 12y + 8 = 0$ are

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