Equation of circle whose centre is $(3, - 1)$ and which cut off length $6$ unit from the line $2 x - 5 y + 18 = 0$ is ?

x^{2} + y^{2} - 6 x + 2 y - 28 = 0

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x^{2} + y^{2} - 6 x + 2 y - 18 = 0

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x^{2} + y^{2} - 6 x + 2 y - 11 = 0

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x^{2} + y^{2} - 6 x + 2 y - 27 = 0

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Solution

The correct option is D $x^{2} + y^{2} - 6 x + 2 y - 28 = 0$
$t h e l e n g t h p e r p e n d i c u l a r f o r m (3 - 1) t o 2 x - 5 y + 18 = 0$
$\frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{2^{2} + 5^{2}}} = \frac{29}{\sqrt{29}} = \sqrt{29}$
Now as circle cuts off an intercept of length $6$ , it intercepts at a distance of $3$ on each side from the foot of the perpendicular.
$\sqrt{({\sqrt{29}}^{2}) + 3^{2}} = \sqrt{38}$
${(x - 3)}^{2} + ({y + 1}^{2})$
$x^{2} + y^{2} - 6 x + 2 y - 28 = 0$

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