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