Equation of circles touching x-axis at the origin and the line $4 x - 3 y + 24 = 0$ are

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$x^{2} + y^{2} + 2 y = 0, x^{2} + y^{2} - 18 y = 0$

$x^{2} + y^{2} - 6 y = 0, x^{2} + y^{2} + 24 y = 0$

$x^{2} + y^{2} + 18 x = 0, x^{2} + y^{2} - 8 x = 0$

$x^{2} + y^{2} + 4 x = 0, x^{2} + y^{2} - 16 x = 0$

answer is A.

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Detailed Solution

Given line is $4 x - 3 y + 24 = 0$

Since equation of circles touching x-axis at the origin then

Centre = $(0, r)$

radius, $r = d$ =The perpendicular distance from centre to the line

$\Rightarrow r = \frac{| 0 - 3 r + 24 |}{5}$

$∴ r = 12, r = 3$

$∴$ The equation of the circle is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

${(x - 0)}^{2} + {(y - 3)}^{2} = 9$ or

${(x - 0)}^{2} + {(y + 12)}^{2} = 144$

$\Rightarrow x^{2} + y^{2} - 6 y = 0, x^{2} + y^{2} + 24 y = 0$

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