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(C) Let the center of the circle be $(h, k)$ and radius be $r$.Since the circle touches the $Y$-axis at $(0, 4)$,the distance from the center to the $

Vedclass · Accepted Answer

$(x \pm 5)^{2} + (y \pm 4)^{2} = 25$

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(C) Let the center of the circle be $(h, k)$ and radius be $r$.Since the circle touches the $Y$-axis at $(0, 4)$,the distance from the center to the $Y$-axis is the radius,so $|h| = r$.Given the circle touches at $(0, 4)$,the $y$-coordinate of the center must be $k = 4$ or $k = -4$.Thus,the center is $(\pm r, \pm 4)$.The equation of the circle is $(x \mp r)^{2} + (y \mp 4)^{2} = r^{2}$.Since it cuts an intercept of $6$ units on the $X$-axis,the length of the intercept is $2\sqrt{r^{2} - k^{2}} = 6$.$\sqrt{r^{2} - 4^{2}} = 3 \implies r^{2} - 16 = 9 \implies r^{2} = 25 \implies r = 5$.Substituting $r=5$ and $k=\pm 4$,the center is $(\pm 5, \pm 4)$.The equation of the circle is $(x \pm 5)^{2} + (y \pm 4)^{2} = 25$.

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

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