Verified Solution • 2025–26 Session moderate

QJEEM24Mab9ba: JEE Main 2024 Mathematics - Class 11 JEEM24Mab9ba Solution: Let the circles and touch each other externally at the point .…

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

Multiple Choice Options

  • A 130
  • B 110
  • C 145
  • D 125

Step-by-Step Verified Solution

amp;C1→(x−α)2+(y−β)2=r12amp;C2→(x−8)2+(y−152)2=r22\begin{aligned}& C_1 \rightarrow(x-\alpha)^2+(y-\beta)^2=r_1^2 \\& C_2 \rightarrow(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2\end{aligned}

Point PP divide the line segment internally C1C2C_1 C_2 in the ratio 2 : 1

JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Circle Question 12 English Explanation

amp;α×1+8×21+2=6,α=2amp;β×1+15α×21+2=6,β=3amp;r1=(6−2)2+(6−3)2=25=5amp;r2=(8−6)2+(15α−6)2=52amp;α+β+4(r12+r22)=2+3+4(52+(52)2)amp;=5+4(1254)amp;=130\begin{aligned}& \frac{\alpha \times 1+8 \times 2}{1+2}=6, \alpha=2 \\& \frac{\beta \times 1+\frac{15}{\alpha} \times 2}{1+2}=6, \beta=3 \\& r_1=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{25}=5 \\& r_2=\sqrt{(8-6)^2+\left(\frac{15}{\alpha}-6\right)^2}=\frac{5}{2} \\& \alpha+\beta+4\left(r_1^2+r_2^2\right)=2+3+4\left(5^2+\left(\frac{5}{2}\right)^2\right) \\& =5+4\left(\frac{125}{4}\right) \\& =130 \\\end{aligned}

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