MHT CET · Maths · Circle
The centre of the circle whose radius is 3 units and touching internally the circle \(x^2+y^2-4 x-6 y-12=0\) at the point \((-1,-1)\) is
- A \(\left(\frac{4}{5}, \frac{7}{5}\right)\)
- B \(\left(\frac{4}{5}, \frac{-7}{5}\right)\)
- C \(\left(\frac{-4}{5}, \frac{-7}{5}\right)\)
- D \(\left(\frac{-4}{5}, \frac{7}{5}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{4}{5}, \frac{7}{5}\right)\)
Step-by-step Solution
Detailed explanation

\(\begin{aligned}
\mathrm{PC}_1 & =\sqrt{(2+1)^2+(3+1)^2} \\
& =\sqrt{25} \\
& =5
\end{aligned}\)
\(P\) divides \(C_1 C_2\) externally in the ratio \(r_1: r_2\) i.e. \(5: 3\)
\(\therefore \quad-1=\frac{5(\mathrm{~h})-3(2)}{5-3}\) and \(-1=\frac{5(\mathrm{k})-3(3)}{5-3}\)
\(\begin{aligned} & \Rightarrow-2=5 \mathrm{~h}-6 \text { and }-2=5 \mathrm{k}-9 \\ & \Rightarrow \mathrm{h}=\frac{4}{5} \text { and } \mathrm{k}=\frac{7}{5}\end{aligned}\)
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