The equation of the circle whose radius is 3 | vedclass

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(A) The given circle is $S: x^2+y^2-4x-6y-12=0$. The center $C_1$ is $(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let the req

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$5x^2+5y^2-8x-14y-32=0$

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(A) The given circle is $S: x^2+y^2-4x-6y-12=0$. The center $C_1$ is $(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$. Let the required circle have center $C_2(h, k)$ and radius $r_2 = 3$. Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 5 : 3$. Using the section formula for external division: $-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$. $-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$. The equation of the circle is $(x - \frac{4}{5})^2 + (y - \frac{7}{5})^2 = 3^2$. $x^2 - \frac{8}{5}x + \frac{16}{25} + y^2 - \frac{14}{5}y + \frac{49}{25} = 9$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{65}{25} - 9 = 0$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{13}{5} - 9 = 0$. $x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y - \frac{32}{5} = 0$. Multiplying by $5$,we get $5x^2 + 5y^2 - 8x - 14y - 32 = 0$.

The equation of the circle whose radius is $3$ and which touches the circle $x^2+y^2-4x-6y-12=0$ internally at $(-1,-1)$ is

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