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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

The equation of the circle having centre at $(3,-4)$ and touching the line $5x+12y-12=0$ is __________

[Hint : To determine radius find the perpendicular distance from the centre of the circle to the line.]

$\begin{array}{1 1}(x-3)^2+(y+4)^2=45^2\\(x-3)^2+(y+4)^2=\large\frac{\sqrt{45}}{13}\\(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2\\(x-3)^2+(y+4)^2=\sqrt{\large\frac{45}{13}}\end{array} $

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