JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the circles \(C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2\) and \(C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2\) touch each other externally at the point \((6,6)\). If the point \((6,6)\) divides the line segment joining the centres of the circles \(C_1\) and \(C_2\) internally in the ratio \(2: 1\), then \((\alpha+\beta)+4\left(r_1^2+r_2^2\right)\) equals
- A \(110\)
- B \(130\)
- C \(125\)
- D \(145\)
Answer & Solution
Correct Answer
(B) \(130\)
Step-by-step Solution
Detailed explanation
\( \because \frac{16+\alpha}{3}=6 \text { and } \frac{15+\beta}{3}=6 \) \( \Rightarrow(\alpha, \beta) \equiv(2,3) \) \( \text { Also, } \mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2 \)…
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