Coordinate Geometry – Circles – JEE Main 23 Jan 2026 Shift 2

Solution:

The equation of any curve passing through the intersection of the two ellipses is given by $S_1 + \lambda S_2 = 0$:
$$ (x^2 + 2y^2 – 6x – 12y + 23) + \lambda(4x^2 + 2y^2 – 20x – 12y + 35) = 0 $$
$$ (1 + 4\lambda)x^2 + (2 + 2\lambda)y^2 – (6 + 20\lambda)x – (12 + 12\lambda)y + (23 + 35\lambda) = 0 $$

For this curve to be a circle, the coefficient of $x^2$ must equal the coefficient of $y^2$:
$$ 1 + 4\lambda = 2 + 2\lambda $$
$$ 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2} $$

Substituting $\lambda = 1/2$ back into the equation:
$$ (1 + 2)x^2 + (2 + 1)y^2 – (6 + 10)x – (12 + 6)y + (23 + 17.5) = 0 $$
$$ 3x^2 + 3y^2 – 16x – 18y + 40.5 = 0 $$

Divide by 3 to get standard form:
$$ x^2 + y^2 – \frac{16}{3}x – 6y + \frac{27}{2} = 0 $$

Centre $(a, b) = \left( -g, -f \right) = \left( \frac{8}{3}, 3 \right) $.
Radius $r = \sqrt{g^2 + f^2 – c} = \sqrt{\left(\frac{8}{3}\right)^2 + 3^2 – \frac{27}{2}}$.

We need to calculate $ab + 18r^2$:
$$ ab = \left(\frac{8}{3}\right)(3) = 8 $$
$$ 18r^2 = 18 \left( \frac{64}{9} + 9 – \frac{27}{2} \right) = 18\left(\frac{64}{9}\right) + 18(9) – 18\left(\frac{27}{2}\right) $$
$$ = 2(64) + 162 – 9(27) $$
$$ = 128 + 162 – 243 = 47 $$

$$ ab + 18r^2 = 8 + 47 = 55 $$

Ans. (4)