The equations of the two circles which touch | vedclass

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(A) Let the center of the circle be $(h, k)$. Since the circle touches the $Y$-axis at $(0,3)$,the radius $r = |h|$. Thus,the center is $(\pm r, 3)$.T

Vedclass · Accepted Answer

$x^{2}+y^{2} \pm 10x-6y+9=0$

Vedclass · Answer

(A) Let the center of the circle be $(h, k)$. Since the circle touches the $Y$-axis at $(0,3)$,the radius $r = |h|$. Thus,the center is $(\pm r, 3)$.The equation of the circle is $(x \mp r)^{2} + (y-3)^{2} = r^{2}$.Since the circle makes an intercept of $8$ units on the $X$-axis,the length of the intercept is $2\sqrt{g^{2}-c}$ or using the geometry,$2\sqrt{r^{2}-k^{2}} = 8$.Here,$k=3$,so $2\sqrt{r^{2}-3^{2}} = 8 \implies \sqrt{r^{2}-9} = 4 \implies r^{2}-9 = 16 \implies r^{2} = 25 \implies r = 5$.The center is $(\pm 5, 3)$.The equation is $(x \mp 5)^{2} + (y-3)^{2} = 5^{2}$.$x^{2} \mp 10x + 25 + y^{2} - 6y + 9 = 25$.$x^{2} + y^{2} \mp 10x - 6y + 9 = 0$.

The equations of the two circles which touch the $Y$-axis at $(0,3)$ and make an intercept of $8$ units on the $X$-axis are

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