Find the equation of the circle whose centre is (3, −1) and which cut-off an intercept of length 6 from the line 2x − 5y + 18 = 0.

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Solution

OC is the perpendicular distance from (3, −1) to 2x − 5y + 18 = 0.

∴ $O C = |\frac{2 (3) - 5 (- 1) + 18}{\sqrt{4 + 25}}| = |\frac{6 + 5 + 18}{\sqrt{4 + 25}}| = \sqrt{29}$

⇒ Radius = $\sqrt{O C^{2} + O B^{2}} = \sqrt{29 + 9} = \sqrt{38}$

∴ Required equation = ${(x - 3)}^{2} + {(y + 1)}^{2} = 38$
= $x^{2} + y^{2} - 6 x + 2 y = 28$

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