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(A) Let the center of the circle be $(-h, -k)$ where $h, k > 0$ (since it is in the third quadrant).Since the circle touches the $x$-axis,the radius $

Vedclass · Accepted Answer

$9(x^2 + y^2) + 6x + 24y + 1 = 0$

Vedclass · Answer

(A) Let the center of the circle be $(-h, -k)$ where $h, k > 0$ (since it is in the third quadrant).Since the circle touches the $x$-axis,the radius $r = |-k| = k$.The equation of the circle is $(x + h)^2 + (y + k)^2 = k^2$.Since the center $(-h, -k)$ lies on the line $x - y - 1 = 0$,we have $-h - (-k) - 1 = 0$,which implies $k - h = 1$,or $h = k - 1$.The circle also touches the line $4x - 3y + 4 = 0$. The perpendicular distance from the center $(-h, -k)$ to this line is equal to the radius $k$.$\frac{|4(-h) - 3(-k) + 4|}{\sqrt{4^2 + (-3)^2}} = k$$\frac{|-4h + 3k + 4|}{5} = k$Substitute $h = k - 1$:$|-4(k - 1) + 3k + 4| = 5k$$|-4k + 4 + 3k + 4| = 5k$$|-k + 8| = 5k$Case $1$: $-k + 8 = 5k$ $\Rightarrow 6k = 8$ $\Rightarrow k = \frac{4}{3}$. Then $h = \frac{4}{3} - 1 = \frac{1}{3}$.Case $2$: $-k + 8 = -5k$ $\Rightarrow 4k = -8$ $\Rightarrow k = -2$ (Not possible as $k > 0$).Thus,the center is $(-\frac{1}{3}, -\frac{4}{3})$ and $r = \frac{4}{3}$.The equation is $(x + \frac{1}{3})^2 + (y + \frac{4}{3})^2 = (\frac{4}{3})^2$.$x^2 + \frac{2}{3}x + \frac{1}{9} + y^2 + \frac{8}{3}y + \frac{16}{9} = \frac{16}{9}$.$x^2 + y^2 + \frac{2}{3}x + \frac{8}{3}y + \frac{1}{9} = 0$.Multiplying by $9$: $9(x^2 + y^2) + 6x + 24y + 1 = 0$.

Find the equation of a circle which touches the $x$-axis and the line $4x - 3y + 4 = 0$,and whose center lies in the third quadrant on the line $x - y - 1 = 0$.

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