Find the equation of a circle whose centre is at $(4, - 2)$ and $3 x - 4 y + 5 = 0$ is tangent to circle.

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Solution

Given equation of tangent is $3 x - 4 y + 5 = 0$
Centre of circle is $(4, - 2)$
Radius of circle is distance of centre from the tangent
$⟹ r a d i u s = \frac{| 3 (4) - 4 (- 2) + 5 |}{\sqrt{3^{2} + 4^{2}}} = \frac{12 + 8 + 5}{5} = 5$
Equation of circle is $(x - 4)^{2} + (y + 2)^{2} = 25 ⟹ x^{2} + y^{2} - 8 x + 4 y - 5 = 0$

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