Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line $5 x + 12 y - 1 = 0$

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Solution

The centre of the required circle in (3, 4) and the circle touches the line 5x + 12y = 1 so, radius = OA = Perpendicular distance of O to 5x + 12y =1
[ $∵$ radius is perpendicular to the tangent]
$\Rightarrow O A = \frac{5 \times 3 + 12 \times 4 - 1}{\sqrt{5^{+} 12^{2}}} = \frac{62}{13}$
Thus the equation of circle will be,
$(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{13})}^{2} \Rightarrow 169 [x^{2} + y^{2} - 6 x - 8 y] + 25 \times 169 = 3844 \Rightarrow 169 [x^{2} + y^{2} - 6 x - 8 y] + 381 = 0$

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