If a circle S with radius 5 touches the circl | vedclass

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(A) The given circle is $x^2+y^2-6x-4y-12=0$. Completing the square,we get $(x-3)^2+(y-2)^2 = 12+9+4 = 25$. So,the centre of the given circle is $C(3,

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$5 \sqrt{3}$

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(A) The given circle is $x^2+y^2-6x-4y-12=0$. Completing the square,we get $(x-3)^2+(y-2)^2 = 12+9+4 = 25$. So,the centre of the given circle is $C(3, 2)$ and its radius $r_1 = 5$. Let the centre of circle $S$ be $B$ and its radius be $r_2 = 5$. The circles touch at $P(-1, -1)$. The distance between the centres $B$ and $C$ is $BC = r_1 + r_2 = 5 + 5 = 10$. Let $A$ be the point of tangency on the circle $C$ from $B$. Since $BA$ is a tangent to the circle $C$ at $A$,$	riangle BAC$ is a right-angled triangle with $\angle BAC = 90^{\circ}$. In $	riangle BAC$,$BC$ is the hypotenuse,$AC$ is the radius of the given circle $(r_1 = 5)$,and $AB$ is the length of the tangent. By the Pythagorean theorem,$AB^2 + AC^2 = BC^2$. $AB^2 + 5^2 = 10^2$. $AB^2 + 25 = 100$. $AB^2 = 75$. $AB = \sqrt{75} = 5 \sqrt{3}$.

If a circle $S$ with radius $5$ touches the circle $x^2+y^2-6x-4y-12=0$ at $(-1,-1)$,then the length of the tangent from the centre of the circle $S$ to the given circle is

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