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(A) The given circle is $x^2+y^2-4x-6y-12=0$. Its centre $C_1$ is $(2, 3)$ and radius $R = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. Let

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$\left(\frac{4}{5}, \frac{7}{5}\right)$

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(A) The given circle is $x^2+y^2-4x-6y-12=0$. Its centre $C_1$ is $(2, 3)$ and radius $R = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. Let the centre of the required circle be $C_2(h, k)$ and its radius be $r = 3$. Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $R:r = 5:3$. Using the section formula for external division: $P = \left(\frac{R h - r x_1}{R - r}, \frac{R k - r y_1}{R - r}\right)$ $-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$ $-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$ Thus,the centre is $\left(\frac{4}{5}, \frac{7}{5}\right)$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

The centre of the circle whose radius is $3$ units and touching internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1, -1)$ is

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