Find the equation of a circle with radius 3 t | vedclass

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Question

(A) Let the given circle be $S: x^{2} + y^{2} - 4x - 6y - 12 = 0$. Its center $C$ is $(2, 3)$ and radius $R = \sqrt{2^{2} + 3^{2} - (-12)} = \sqrt{4 +

Vedclass · Accepted Answer

$(x - \frac{4}{5})^{2} + (y - \frac{7}{5})^{2} = 3^{2}$

Vedclass · Answer

(A) Let the given circle be $S: x^{2} + y^{2} - 4x - 6y - 12 = 0$. Its center $C$ is $(2, 3)$ and radius $R = \sqrt{2^{2} + 3^{2} - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$. Let the required circle have center $C_1(h, k)$ and radius $r = 3$. Since the circle touches internally at $P(-1, -1)$,the center $C_1$ lies on the line segment $CP$. The distance $CC_1 = R - r = 5 - 3 = 2$. Thus,$C_1$ divides the line segment $CP$ in the ratio $r : (R - r) = 3 : 2$ internally. Using the section formula,$C_1 = (\frac{3(2) + 2(-1)}{3+2}, \frac{3(3) + 2(-1)}{3+2}) = (\frac{6-2}{5}, \frac{9-2}{5}) = (\frac{4}{5}, \frac{7}{5})$. The equation of the circle is $(x - \frac{4}{5})^{2} + (y - \frac{7}{5})^{2} = 3^{2}$.

Find the equation of a circle with radius $3$ that touches the circle $x^{2} + y^{2} - 4x - 6y - 12 = 0$ internally at the point $(-1, -1)$.

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