Equation of the circle having centre at $(3, - 1)$ and making an intercept of length $6$ units on the line $2 x - 5 y + 18 = 0$ , is

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

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x^{2} + y^{2} - 6 x + 2 y - 38 = 0

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

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x^{2} + y^{2} - 6 x + 2 y - 48 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 6 x + 2 y - 28 = 0$

Length of perpendicular from $C (3, - 1)$
$d = \frac{6 + 5 + 18}{\sqrt{4 + 25}} = \sqrt{29}$ units
If the radius of required circle is $r$
then, $A B = 2 \sqrt{r^{2} - d^{2}}$
$\Rightarrow 3 = \sqrt{r^{2} - 29}$
$\Rightarrow r = \sqrt{38}$ units
Thus, equation of circle is
$(x - 3)^{2} + (y + 1)^{2} = 38$
$\Rightarrow x^{2} + y^{2} - 6 x + 2 y - 28 = 0$

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