The equation of the incircle of the triangle formed by the axes and the line $4 x + 3 y = 6$ is

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

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4 (x^{2} + y^{2} - x - y) + 1 = 0

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4 (x^{2} + y^{2} + x + y) + 1 = 0

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none of these

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Solution

The correct option is D $x^{2} + y^{2} - 6 x - 6 y + 9 = 0$
The incircle touches the axes and $4 x + 3 y = 6$

Equation of circle which touches both axes is given by $(x - a)^{2} + (y - a)^{2} = a^{2}$

$4 x + 3 y = 6$ is also a tangent

$⟹$ radius = perpendicular distance from center to tangent

$⟹ a = \frac{| 4 a + 3 a - 6 |}{\sqrt{4^{2} + 3^{2}}}$

$⟹ 5 a = 7 a - 6$

$⟹ 2 a = 6 ⟹ a = 3$

Equation of circle is $(x - 3)^{2} + (y - 3)^{2} = 3^{2}$

$⟹ x^{2} + y^{2} - 6 x - 6 y + 9 = 0$

Correct answer is $x^{2} + y^{2} - 6 x - 6 y + 9 = 0$

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