JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Circle

Let the circles

C_{1} : (x - α)^{2} + (y - β)^{2} = r_{1}^{2}

and

C_{2} : (x - 8)^{2} + (y - \frac{15}{2})^{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

(α + β) + 4 (r_{1}^{2} + r_{2}^{2})

equals

(A)110
(B)130✓ Correct
(C)125
(D)145

Answer: (B)

Solution

The point

(6, 6)

divides

C_{1} C_{2}

internally

2 : 1

, and

C_{2} = (8, \frac{15}{2})

. From

\frac{16 + α}{3} = 6, \frac{15 + β}{3} = 6

we get

(α, β) = (2, 3)

. Distance

C_{1} C_{2} = (8 - 2)^{2} + (7.5 - 3)^{2} = 7.5 = r_{1} + r_{2}

, and the

2 : 1

split gives

r_{1} = 5, r_{2} = 2.5

. Hence

(α + β) + 4 (r_{1}^{2} + r_{2}^{2}) = 5 + 4 (25 + 6.25) = 5 + 125 = 130

. Option (2).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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