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

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Solution

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

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