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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Circle

Let the circles C_1:(x-α)^2+(y-β)^2=r_1^2 and...

Let the circles C1:(x−α)2+(y−β)2=r12C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2 and C2:(x−8)2+(y−152)2=r22C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2 touch each other externally at the point (6,6)(6,6). If the point (6,6)(6,6) divides the line segment joining the centres of the circles C1C_1 and C2C_2 internally in the ratio 2:12: 1, then (α+β)+4(r12+r22)(\alpha+\beta)+4\left(r_1^2+r_2^2\right) equals

Options

  1. A125
  2. B130
  3. C110
  4. D145

Correct answer

B. 130

Step-by-step solution

∵16+α3=6 and  15+β3=6⇒(α,β)≡(2,3)\begin{aligned} & \because \frac{16+\alpha}{3}=6 \text{\text{ and } } \frac{15+\beta}{3}=6 \\ & \Rightarrow(\alpha, \beta) \equiv(2,3)\end{aligned} Also, C1C2=r1+r2\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2⇒(2−8)2+(3−152)2=2r2+r2⇒r2=52⇒r1=2r2=5∴(α+β)+4(r12+r22)=5+4(254+25)=130\begin{aligned} & \Rightarrow \sqrt{(2-8)^2+\left(3-\frac{15}{2}\right)^2}=2 r_2+r_2 \\ & \Rightarrow r_2=\frac{5}{2} \Rightarrow r_1=2 r_2=5 \\ & \therefore(\alpha+\beta)+4\left(r_1^2+r_2^2\right) \\ & =5+4\left(\frac{25}{4}+25\right)=130\end{aligned}$$
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