• # question_answer The equation of the pair of straight lines parallel to x-axis and touching the circle ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$ [Kerala (Engg.) 2002] A)            ${{y}^{2}}-4y-21=0$         B)            ${{y}^{2}}+4y-21=0$ C)            ${{y}^{2}}-4y+21=0$        D)            ${{y}^{2}}+4y+21=0$

Correct Answer: A

Solution :

Let the lines are $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$ since pair of straight lines parallel to x-axis,   \${{m}_{1}}={{m}_{2}}=0$            and the lines will be $y={{c}_{1}}$and $y={{c}_{2}}$            Given circle is ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$, centre (3, 2) and radius = 5.            Here, the perpendicular drawn from centre to the lines are CP and $C{P}'$.            $CP=\frac{2-{{c}_{1}}}{\sqrt{1}}=\pm 5$ Þ $2-{{c}_{1}}=\pm 5$            ${{c}_{1}}=7$ and ${{c}_{1}}=-3$            Hence the lines are            $y-7=0,\,y+3=0$ i.e.,$(y-7)\,(y+3)=0$            \ Pair of straight lines is ${{y}^{2}}-4y-21=0$.

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