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Question

Find the equation to the circle which has its centre at the point $$(3, 4)$$ and touches the straight line $$5x + 12y = 1$$.


Solution

Perpendicular distance of the centre from the line  will be equal to the radius as the line is tangent to the circle
$$\therefore d=\cfrac{5(3)+12(4)-1}{13}=r$$
$$d=r=\cfrac{62}{13}$$
So, the equation of circle will be
$$(x-3)^{2}+(y-4)^{2}=\left({\cfrac{62}{13}}\right)^{2}$$
$$x^{2}+y^{2}-6x-8y+\cfrac{381}{169}=0$$

Maths

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