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JEE Main 8 Apr 2024 Shift 1 Paper

Section: Mathematics

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Question : 3 of 90

Marks: +1, -0

Let the circles

C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2

C_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)

divides the line segment joining the centres of the circles

C_1

C_2

internally in the ratio

2: 1

(\alpha+\beta)+4 (r_1^2+r_2^2)

[8 Apr 2024 Shift 1]

110
130
125
145

Solution: 👈: Video Solution

\; \because \; \frac{16+\alpha}{3}=6 \; \text{ and } \; \frac{15+\beta}{3}=6

\; \Rightarrow (\alpha, \beta) \equiv (2,3)

\; \; \text{ Also, } \; C_1 C_2=r_1+r_2

\; \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 (r_1^2+r_2^2)

\; =5+4 \left(\frac{25}{4}+25\right)=130

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