The equation of the pair of straight lines pa | vedclass

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Question

(A) Given the circle equation: $x^2+y^2-6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$. Centre of the circle: $(-g, -f)

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

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(A) Given the circle equation: $x^2+y^2-6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$. Centre of the circle: $(-g, -f) = (3, 2)$. Radius of the circle: $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(-2)^2-(-12)} = \sqrt{9+4+12} = \sqrt{25} = 5$. Since the lines are parallel to the $x$-axis,they are of the form $y = k$. These lines touch the circle,so the distance from the centre $(3, 2)$ to the line $y = k$ must be equal to the radius $r = 5$. Therefore,$|k - 2| = 5$. This gives $k - 2 = 5$ or $k - 2 = -5$. So,$k = 7$ or $k = -3$. The equations of the lines are $y = 7$ and $y = -3$. The pair of straight lines is given by $(y - 7)(y + 3) = 0$. Expanding this,we get $y^2 + 3y - 7y - 21 = 0$,which simplifies to $y^2 - 4y - 21 = 0$.

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

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