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(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. \ Since the circle touches the $x$-axis at $(3, 0)$,the center is $(h, k)$ and the r

Vedclass · Accepted Answer

$x^2+y^2-6x+4y+9=0$

Vedclass · Answer

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. \ Since the circle touches the $x$-axis at $(3, 0)$,the center is $(h, k)$ and the radius $r = |k|$. \ Thus,the equation becomes $(x-h)^2 + (y-k)^2 = k^2$. \ Since it passes through $(3, 0)$,we have $(3-h)^2 + (0-k)^2 = k^2$ $\Rightarrow (3-h)^2 = 0$ $\Rightarrow h = 3$. \ Since it also passes through $(1, -2)$,we substitute $h=3$ into the equation: $(1-3)^2 + (-2-k)^2 = k^2$. \ $(-2)^2 + 4 + 4k + k^2 = k^2$ \ $4 + 4 + 4k = 0$ $\Rightarrow 4k = -8$ $\Rightarrow k = -2$. \ Substituting $h=3$ and $k=-2$ into $(x-h)^2 + (y-k)^2 = k^2$: \ $(x-3)^2 + (y+2)^2 = (-2)^2$ \ $x^2 - 6x + 9 + y^2 + 4y + 4 = 4$ \ $x^2 + y^2 - 6x + 4y + 9 = 0$.

Find the equation of the circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$.

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