The equation of a circle touching the coordin | vedclass

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(C) Since the circle touches both coordinate axes,its center is $(h, h)$ and its radius is $|h|$.Given that the circle also touches the line $3x - 4y

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 6y + 9 = 0$

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(C) Since the circle touches both coordinate axes,its center is $(h, h)$ and its radius is $|h|$.Given that the circle also touches the line $3x - 4y - 12 = 0$,the perpendicular distance from the center $(h, h)$ to the line must equal the radius $|h|$.$	herefore \left|\frac{3h - 4h - 12}{\sqrt{3^2 + (-4)^2}}ight| = |h|$$\Rightarrow \left|\frac{-h - 12}{5}ight| = |h|$$\Rightarrow |-h - 12| = 5|h|$Case $1$: $-h - 12 = 5h$ $\Rightarrow 6h = -12$ $\Rightarrow h = -2$.Case $2$: $-h - 12 = -5h$ $\Rightarrow 4h = 12$ $\Rightarrow h = 3$.For $h = 3$,the equation is $(x - 3)^2 + (y - 3)^2 = 3^2$.$x^2 - 6x + 9 + y^2 - 6y + 9 = 9$$x^2 + y^2 - 6x - 6y + 9 = 0$.

The equation of a circle touching the coordinate axes and the line $3x - 4y = 12$ is

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