The equation of the circle of radius $4$ and center lies in the first quadrant which touches the x-axis and the line $4 x - 3 y = 0$ is-

(x - 8)^{2} + (y - 4)^{2} = 16

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x^{2} + y^{2} + 30 x - 10 y + 225 = 0

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x^{2} + y^{2} - 30 x + 10 y + 225 = 0

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None of these

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Solution

The correct option is A $(x - 8)^{2} + (y - 4)^{2} = 16$
Given,
$r = 4$
The centre will be,
$c (p, 4)$
We know that the distance
b/w a line and a point is
$r = 4 = | \frac{4 \times p - 3 \times 4 + 0}{\sqrt{4^{2} + 3^{2}}} |$
$4 = | \frac{4 p - 12}{\sqrt{25} - 5} |$
$20 = | 4 p - 12 |$
$4p-12=20
$4 p = 32$
$p = 8$
$4 p - 12 = - 20$
$4 p = - 8$
$p = - 2$
since $p > 0$
So the centre is $(8, 4)$
The equation circle will be
$(x - 8)^{2} + (y - 4)^{2} = 4^{2} = 16$

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