The equation of the circle that touches the Y | vedclass

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Question

(A) Let the center of the circle be $C(h, k)$.Since the circle touches the $Y-$axis at a distance of $4$ units from the origin,the center is $C(\pm r,

Vedclass · Accepted Answer

$x^2+y^2 \pm 10 x-8 y+16=0$

Vedclass · Answer

(A) Let the center of the circle be $C(h, k)$.Since the circle touches the $Y-$axis at a distance of $4$ units from the origin,the center is $C(\pm r, 4)$,where $r$ is the radius.Thus,the equation of the circle is $(x \mp r)^2 + (y - 4)^2 = r^2$.This circle cuts an intercept of $6$ units on the $X-$axis. The length of the intercept on the $X-$axis is $2\sqrt{g^2 - c}$ or $2\sqrt{r^2 - k^2}$ where $k=4$.Given $2\sqrt{r^2 - 4^2} = 6$,so $\sqrt{r^2 - 16} = 3$.Squaring both sides,$r^2 - 16 = 9$,which gives $r^2 = 25$,so $r = 5$.The center is $C(\pm 5, 4)$.The equation is $(x \mp 5)^2 + (y - 4)^2 = 25$.Expanding this,$x^2 \mp 10x + 25 + y^2 - 8y + 16 = 25$.$x^2 + y^2 \mp 10x - 8y + 16 = 0$.

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

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