Find the equation of the pair of straight lin | vedclass

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(A) The given circle is $x^2 + y^2 - 6x - 4y - 12 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = -2, c = -12$.The center of t

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$y^2 - 4y - 21 = 0$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 6x - 4y - 12 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = -2, c = -12$.The center of the circle is $(-g, -f) = (3, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-2)^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$.Since the lines are parallel to the $x$-axis,their equations are of the form $y = k$.The distance from the center $(3, 2)$ to the line $y - k = 0$ must be equal to the radius $r = 5$.Using the perpendicular distance formula,$\frac{|2 - k|}{\sqrt{0^2 + 1^2}} = 5$.$|2 - k| = 5$.This gives $2 - k = 5$ or $2 - k = -5$.So,$k = -3$ or $k = 7$.The lines are $y = -3$ and $y = 7$,which can be written as $y + 3 = 0$ and $y - 7 = 0$.The equation of the pair of lines is $(y + 3)(y - 7) = 0$.$y^2 - 7y + 3y - 21 = 0$.$y^2 - 4y - 21 = 0$.

Find the equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2 + y^2 - 6x - 4y - 12 = 0$.

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