Question

Points of intersection of ellipses $x^2 + 2y^2 - 6x - 12y + 23 = 0$ and $4x^2 + 2y^2 - 20x - 12y + 35 = 0$ lie on a circle. Value of $ab + 18r^2$ is

• A. 53
• B. 51
• C. 55
• D. 52

Hint:

Family of curves $S_1 + \lambda S_2 = 0$ is powerful for finding curves through intersections without finding the points explicitly.

Answer 1

The Correct Option is C

Consider the family of curves passing through intersection: $S_2 + \lambda S_1 = 0$.
$(4x^2 + 2y^2 - 20x - 12y + 35) + \lambda(x^2 + 2y^2 - 6x - 12y + 23) = 0$.
$(4+\lambda)x^2 + (2+2\lambda)y^2 + \dots = 0$.
For a circle, coeff of $x^2$ = coeff of $y^2$.
$4+\lambda = 2+2\lambda \implies \lambda = 2$.
Substitute $\lambda=2$: $6x^2 + 6y^2 - 32x - 36y + 81 = 0$.
Divide by 6: $x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0$.
Center $(a,b) = (8/3, 3)$.
Radius squared $r^2 = g^2 + f^2 - c = \frac{64}{9} + 9 - \frac{27}{2}$.
Value $= ab + 18r^2 = (8/3)(3) + 18(\frac{64}{9} + 9 - \frac{27}{2})$.
$= 8 + 128 + 162 - 243 = 55$.