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

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Aicharya · Accepted Answer

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$\begin{aligned} & (3,0)=\left(\frac{-6 g-9}{6+3}, \frac{-6 f+12}{6+3}\right) \\ & 3=\frac{-6 g-9}{6+3} \Rightarrow g=-6 \text { and } O=\frac{-6 f+12}{6+3} \Rightarrow f=2 \\ & \text { Given that } \sqrt{g^2+f^2-c}=3 \\ & \Rightarrow 36+4-c=9 \Rightarrow c=31 \end{aligned}$ Now, $3 g-4 f+c=-18-8+31=5$

If the equation of the circle of radius 3 units which touches the circle $x^2+y^2+6 x-8 y-11=0$ externally…

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