The equation of the pair of straight lines pa | vedclass

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Question

(A) Given circle is $x^2+y^2-6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$. Centre $(h, k) = (-g, -f) = (3, 2)$. Radius

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

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

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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