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If the straight lines $3 x-4 y+4=0$ and $6 x-8 y-7=0$ are the tangents to the same circle, then the area of that circle (in square units) is
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The correct answer is:
$\frac{9 \pi}{16}$
Straight line $3 x-4 y+4=0$
and $\quad 6 x-8 y-7=0$
are tangent of same circle
$\therefore \quad 3 x-4 y=-4 \text { and } 3 x-4 y=\frac{7}{2}$
Tangent are parallels
$\therefore$ Distance between tangent are diameter of circle
$\therefore \quad d=\left|\frac{7 / 2+4}{\sqrt{3^2+4^2}}\right|=\left|\frac{3}{2}\right| \Rightarrow r=\frac{d}{2}=\frac{3}{4}$
Area of circle $=\pi r^2=\pi\left(\frac{3}{4}\right)^2=\frac{9 \pi}{16}$
and $\quad 6 x-8 y-7=0$
are tangent of same circle
$\therefore \quad 3 x-4 y=-4 \text { and } 3 x-4 y=\frac{7}{2}$
Tangent are parallels
$\therefore$ Distance between tangent are diameter of circle
$\therefore \quad d=\left|\frac{7 / 2+4}{\sqrt{3^2+4^2}}\right|=\left|\frac{3}{2}\right| \Rightarrow r=\frac{d}{2}=\frac{3}{4}$
Area of circle $=\pi r^2=\pi\left(\frac{3}{4}\right)^2=\frac{9 \pi}{16}$
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