A circle touches the X-axis and the line 4x-3y+4=0 both. If the centre lies on the line x-y-q=0 in the third quadrant, then the equation of the circle is

(x^{2} + y^{2})

+24x+3y-16=0

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(x^{2} + y^{2})

+8x+2y+1=0

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(x^{2} + y^{2})

+6x+24y+1=0

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(x^{2} + y^{2})

+2x+8y+1=0

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Solution

The correct option is A 9 $(x^{2} + y^{2})$ +24x+3y-16=0
Circle touches x-axis, $y = 0$ and line $4 x - 3 y + 4 = 0$
Equation of circle can be written as $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Tangent to the circle $(x_{1}, y_{1})$ can be written as
$x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 y = 0 x - 3 y + 4 = 0 y = x - 9 x = 0, y = - 9 x = 9 y = 0 (9, 0) 9 x + g (x + a) + f y + c = 0 9 x + g x + g q + f y + c = 0 x (9 + g) + f y + c + g q = 0 9 + g = 4 f = - 3 x - y - 9 = 0 - g + f - 9 = 0 - g - q = - f - g - q = 3$
Solving them, we will get $g = - 12$
Equation of circle
$9 (x^{2} + y^{2}) + 24 x + 3 y - 16 = 0$

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