A circle touching the x- axis at (3, 0) and m | vedclass

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(C) Let the center of the circle be $(3, r)$ or $(3, -r)$ and the radius be $r$.Since the circle touches the $x-$ axis at $(3, 0)$,its equation is $(x

Vedclass · Accepted Answer

$(3, 10)$

Vedclass · Answer

(C) Let the center of the circle be $(3, r)$ or $(3, -r)$ and the radius be $r$.Since the circle touches the $x-$ axis at $(3, 0)$,its equation is $(x - 3)^2 + (y - r)^2 = r^2$ or $(x - 3)^2 + (y + r)^2 = r^2$.Expanding this,we get $x^2 - 6x + 9 + y^2 \mp 2ry + r^2 = r^2$,which simplifies to $x^2 + y^2 - 6x \mp 2ry + 9 = 0$.The $y-$ intercept is given by $2\sqrt{f^2 - c}$,where $f = \mp r$ and $c = 9$.Given the intercept length is $8$,we have $2\sqrt{r^2 - 9} = 8$.$\sqrt{r^2 - 9} = 4 \implies r^2 - 9 = 16 \implies r^2 = 25 \implies r = 5$.The equations of the circles are $x^2 + y^2 - 6x + 10y + 9 = 0$ and $x^2 + y^2 - 6x - 10y + 9 = 0$.Checking the point $(3, 10)$ in the second equation: $(3)^2 + (10)^2 - 6(3) - 10(10) + 9 = 9 + 100 - 18 - 100 + 9 = 0$.Thus,the circle passes through $(3, 10)$.

$A$ circle touching the $x-$ axis at $(3, 0)$ and making an intercept of length $8$ on the $y-$ axis passes through the point:

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