The equation of the circle whose radius is 3 and which touches internally the circle \(x^2+y^2-4 x-6…
The equation of the circle whose radius is 3 and which touches internally the circle \(x^2+y^2-4 x-6 y-12=0\) at the point \((-1,-1)\) is
\(5 x^2+5 y^2+9 x-6 y-7=0\)
\(5 x^2+5 y^2-8 x-14 y-32=0\)
\(5 x^2+5 y^2-6 x+8 y-8=0\)
\(5 x^2+5 y^2+6 x-8 y-12=0\)
Solution
Equation of given circle is \(x^2+y^2-4 x-6 y-12=0\) having centre \(C_1(2,3)\) and radius \(r_1=\sqrt{4+9+12}=5\).
Let the required circle having centre \(C_2(h, k)\) and radius is given as 3 touches the given circle at \(A(-1,-1)\).
The point \(A(-1,-1)\) divides the line joining the centres \(C_1(2,3)\) and \(C_2(h, k)\) externally in \(5: 3\) so
\(\begin{aligned}
& (-1,-1)=\left(\frac{5 h-3(2)}{5-3}, \frac{5 k-3(3)}{5-3}\right) \\
& \Rightarrow(-1,-1)=\left(\frac{5 h-6}{2}, \frac{5 k-9}{2}\right) \\
& \Rightarrow 5 h-6=-2 \text { and } 5 k-9=-2 \\
& \Rightarrow h=\frac{4}{5} \text { and } k=\frac{7}{5}
\end{aligned}\)
so equation of required circle is
\(\begin{aligned}
& \left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\
\Rightarrow & x^2-\frac{8 x}{5}+\frac{16}{25}+y^2-\frac{14 y}{5}+\frac{49}{25}=9 \\
\Rightarrow & x^2+y^2-\frac{8 x}{5}-\frac{14 y}{5}+\frac{65}{25}=9 \\
\Rightarrow & x^2+y^2-\frac{8 x}{5}-\frac{14 y}{5}+\frac{13}{5}=9 \\
\Rightarrow & 5 x^2+5 y^2-8 x-14 y-32=0
\end{aligned}\)
Hence, option (2) is correct.