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If the line 3x - 2y + 6 = 0 meets x-axis and y-axis respectively at A & B then the equation of the circle with radius AB and center at ‘A’ is

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$x^{2} + y^{2} + 4 x + 9 = 0$

$x^{2} + y^{2} + 4 x - 9 = 0$

$x^{2} + y^{2} + 4 x + 4 = 0$

$x^{2} + y^{2} + 4 x - 4 = 0$

answer is B.

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Detailed Solution

$\frac{x}{(- 2)} + \frac{y}{3} = 1 \to (1)$

A = (-2, 0), B = (0, 3)

$r = \bar{A B} = \sqrt{4 + 9} = \sqrt{13}$

C = A = (-2, 0)
Equation of circle is

$\begin{array}{l} {(x + 2)}^{2} + {(y - 0)}^{2} = 13 \\ \Rightarrow x^{2} + y^{2} + 4 x - 9 = 0 \end{array}$

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