The equation of a circle which touches both a | vedclass

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(C) Since the circle lies in the third quadrant and touches both axes,its centre is $(-a, -a)$ and its radius is $a$,where $a > 0$.The equation of the

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$x^2 + y^2 + 4x + 4y + 4 = 0$

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(C) Since the circle lies in the third quadrant and touches both axes,its centre is $(-a, -a)$ and its radius is $a$,where $a > 0$.The equation of the circle is $(x + a)^2 + (y + a)^2 = a^2$,which simplifies to $x^2 + y^2 + 2ax + 2ay + a^2 = 0$.The circle touches the line $3x - 4y + 8 = 0$. The perpendicular distance from the centre $(-a, -a)$ to the line must equal the radius $a$:$\left| \frac{3(-a) - 4(-a) + 8}{\sqrt{3^2 + (-4)^2}} \right| = a$$\left| \frac{-3a + 4a + 8}{5} \right| = a$$\left| \frac{a + 8}{5} \right| = a$Case $1$: $a + 8 = 5a$ $\Rightarrow 4a = 8$ $\Rightarrow a = 2$.Case $2$: $a + 8 = -5a$ $\Rightarrow 6a = -8$ $\Rightarrow a = -4/3$ (rejected as $a > 0$).Substituting $a = 2$ into the equation $x^2 + y^2 + 2ax + 2ay + a^2 = 0$ gives:$x^2 + y^2 + 4x + 4y + 4 = 0$.

The equation of a circle which touches both axes and the line $3x - 4y + 8 = 0$ and whose centre lies in the third quadrant is

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