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MCQMediumJEE 2026Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2026 Question with Solution

Points of intersection of ellipses x2+2y2−6x−12y+23=0x^2 + 2y^2 - 6x - 12y + 23 = 0 and 4x2+2y2−20x−12y+35=04x^2 + 2y^2 - 20x - 12y + 35 = 0 lie on a circle. Value of ab+18r2ab + 18r^2 is

  • A

    5353

  • B

    5151

  • C

    5555

  • D

    5252

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The two ellipses are

x2+2y2−6x−12y+23=0x^2 + 2y^2 - 6x - 12y + 23 = 0

and

4x2+2y2−20x−12y+35=04x^2 + 2y^2 - 20x - 12y + 35 = 0

Find: The value of ab+18r2ab + 18r^2 for the circle passing through their points of intersection.

Consider the family of curves through the points of intersection:

S2+λS1=0S_2 + \lambda S_1 = 0

So,

(4x2+2y2−20x−12y+35)+λ(x2+2y2−6x−12y+23)=0(4x^2 + 2y^2 - 20x - 12y + 35) + \lambda(x^2 + 2y^2 - 6x - 12y + 23) = 0

Collecting coefficients gives

(4+λ)x2+(2+2λ)y2+⋯=0(4+\lambda)x^2 + (2+2\lambda)y^2 + \dots = 0

For this curve to be a circle, coefficient of x2x^2 must equal coefficient of y2y^2. Hence,

4+λ=2+2λ4 + \lambda = 2 + 2\lambda

which gives

λ=2\lambda = 2

Substituting λ=2\lambda = 2,

6x2+6y2−32x−36y+81=06x^2 + 6y^2 - 32x - 36y + 81 = 0

Dividing by 66,

x2+y2−163x−6y+272=0x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0

Comparing with the standard form

x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

we get

2g=−163,2f=−6,c=2722g = -\frac{16}{3}, \quad 2f = -6, \quad c = \frac{27}{2}

So the center is

(a,b)=(−g,−f)=(83,3)(a,b) = (-g,-f) = \left(\frac{8}{3}, 3\right)

Now,

r2=g2+f2−c=649+9−272r^2 = g^2 + f^2 - c = \frac{64}{9} + 9 - \frac{27}{2}

Then

ab+18r2=(83)(3)+18(649+9−272)ab + 18r^2 = \left(\frac{8}{3}\right)(3) + 18\left(\frac{64}{9} + 9 - \frac{27}{2}\right) =8+128+162−243= 8 + 128 + 162 - 243 =55= 55

Therefore, the correct option is C.

Equal coefficients trick

Given: Two conics are provided and we need the circle through their intersection points.

Find: ab+18r2ab + 18r^2.

A quick method is to form

S2+λS1=0S_2 + \lambda S_1 = 0

and immediately force the coefficients of x2x^2 and y2y^2 to be equal, because that is the defining condition for a circle in a second-degree equation.

Thus,

4+λ=2+2λ  ⟹  λ=24 + \lambda = 2 + 2\lambda \implies \lambda = 2

So the required circle is

6x2+6y2−32x−36y+81=06x^2 + 6y^2 - 32x - 36y + 81 = 0

or

x2+y2−163x−6y+272=0x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0

Hence,

(a,b)=(83,3),r2=649+9−272(a,b) = \left(\frac{8}{3}, 3\right), \qquad r^2 = \frac{64}{9} + 9 - \frac{27}{2}

Therefore,

ab+18r2=55ab + 18r^2 = 55

So the correct option is C.

Common mistakes

  • Taking the family as a general linear combination but not enforcing equal coefficients of x2x^2 and y2y^2 for a circle is incorrect. For a second-degree curve to represent a circle, these coefficients must match. Always use that condition first.

  • Using the center directly from the coefficients as (−163,−6)\left(-\frac{16}{3}, -6\right) is wrong because the standard form is x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0. First identify 2g2g and 2f2f, then take center as (−g,−f)(-g,-f).

  • Computing the radius with the wrong sign, such as r2=c−g2−f2r^2 = c - g^2 - f^2, gives an incorrect value. For the standard circle form, the correct relation is r2=g2+f2−cr^2 = g^2 + f^2 - c.

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