If the equation of the circle of radius 3 uni | vedclass

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(B) The given circle is $x^2+y^2+6x-8y-11=0$. Its center is $C_1 = (-3, 4)$ and radius $r_1 = \sqrt{(-3)^2 + 4^2 - (-11)} = \sqrt{9+16+11} = \sqrt{36}

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$5$

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(B) The given circle is $x^2+y^2+6x-8y-11=0$. Its center is $C_1 = (-3, 4)$ and radius $r_1 = \sqrt{(-3)^2 + 4^2 - (-11)} = \sqrt{9+16+11} = \sqrt{36} = 6$.Let the required circle be $x^2+y^2+2gx+2fy+c=0$ with center $C_2 = (-g, -f)$ and radius $r_2 = 3$.Since the circles touch externally at $(3,0)$,the point $(3,0)$ divides the line segment joining the centers $C_1(-3, 4)$ and $C_2(-g, -f)$ in the ratio $r_1:r_2 = 6:3 = 2:1$ internally.Using the section formula:$3 = \frac{2(-g) + 1(-3)}{2+1}$ $\Rightarrow 3 = \frac{-2g-3}{3}$ $\Rightarrow 9 = -2g-3$ $\Rightarrow 2g = -12$ $\Rightarrow g = -6$.$0 = \frac{2(-f) + 1(4)}{2+1}$ $\Rightarrow 0 = \frac{-2f+4}{3}$ $\Rightarrow 0 = -2f+4$ $\Rightarrow 2f = 4$ $\Rightarrow f = 2$.The radius of the second circle is $r_2 = \sqrt{g^2+f^2-c} = 3$.Substituting $g=-6$ and $f=2$: $\sqrt{(-6)^2 + 2^2 - c} = 3$ $\Rightarrow 36+4-c = 9$ $\Rightarrow 40-c = 9$ $\Rightarrow c = 31$.Finally,$3g-4f+c = 3(-6) - 4(2) + 31 = -18 - 8 + 31 = 5$.

If the equation of the circle of radius $3$ units which touches the circle $x^2+y^2+6x-8y-11=0$ externally at $(3,0)$ is $x^2+y^2+2gx+2fy+c=0$,then $3g-4f+c=$

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