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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Circle

If the points of intersection of the ellipses x^ 2 +2 y^ 2 -6 x-12...

If the points of intersection of the ellipses x2+2y2−6x−12y+23=0x^{2}+2 y^{2}-6 x-12 y+23=0 and 4x2+2y2−20x−12y+35=04 x^{2}+2 y^{2}-20 x-12 y+35=0 lie on a circle of radius rr and centre (a,b)(a, b), then the value of ab+18r2a b+18 r^{2} is

Options

  1. A52
  2. B53
  3. C51
  4. D55

Correct answer

D. 55

Step-by-step solution

The circle through intersection of the ellipses is S1+λS2=0S_1 + \lambda S_2 = 0.
(1+4λ)x2+(2+2λ)y2−(6+20λ)x−(12+12λ)y+(23+35λ)=0(1+4\lambda)x^2 + (2+2\lambda)y^2 - (6+20\lambda)x - (12+12\lambda)y + (23+35\lambda) = 0.
For a circle: 1+4λ=2+2λ⇒λ=121+4\lambda = 2+2\lambda \Rightarrow \lambda = \frac{1}{2}.
Circle: 3x2+3y2−16x−18y+812=0⇒x2+y2−163x−6y+272=03x^2 + 3y^2 - 16x - 18y + \frac{81}{2} = 0 \Rightarrow x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0.
Centre (a,b)=(83,3)(a,b) = \left(\frac{8}{3}, 3\right), r2=649+9−272=4718r^2 = \frac{64}{9} + 9 - \frac{27}{2} = \frac{47}{18}.
ab+18r2=83×3+18×4718=8+47=55ab + 18r^2 = \frac{8}{3} \times 3 + 18 \times \frac{47}{18} = 8 + 47 = 55.
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