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NVAMediumJEE 2024Circle Equation & Properties

JEE Mathematics 2024 Question with Solution

Consider two circles C1C_1: x2+y2=25x^2 + y^2 = 25 and C2C_2: (x−α)2+y2=16(x-\alpha)^2 + y^2 = 16, where α∈(5,9)\alpha \in (5, 9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C1C_1 and C2C_2 be sin⁡−1(638)\sin^{-1}\left(\frac{\sqrt{63}}{8}\right). If the length of the common chord of C1C_1 and C2C_2 is β\beta, then the value of (αβ)2(\alpha\beta)^2 equals:

Answer

Correct answer:1575

Step-by-step solution

Standard Method

Given: The circles are C1:x2+y2=25C_1: x^2 + y^2 = 25 and C2:(x−α)2+y2=16C_2: (x-\alpha)^2 + y^2 = 16.

So the centers and radii are:

  • For C1C_1, center is (0,0)(0,0) and radius is r1=5r_1 = 5.
  • For C2C_2, center is (α,0)(\alpha,0) and radius is r2=4r_2 = 4.
  • Distance between centers is d=αd = \alpha.

Find: The value of (αβ)2(\alpha\beta)^2, where β\beta is the length of the common chord.

Let the centers be O1O_1 and O2O_2, and let PP be one intersection point of the two circles. Then triangle O1PO2O_1PO_2 has sides:

  • O1P=5O_1P = 5
  • O2P=4O_2P = 4
  • O1O2=αO_1O_2 = \alpha

The given angle between the two radii is

sin⁡θ=638\sin\theta = \frac{\sqrt{63}}{8}

where θ=∠O1PO2\theta = \angle O_1PO_2.

The common chord is perpendicular to the line joining the centers. If the altitude from PP to O1O2O_1O_2 is hh, then

β=2h\beta = 2h

So the area of triangle O1PO2O_1PO_2 using base and height is

Area=12⋅α⋅h=12⋅α⋅β2=αβ4\text{Area} = \frac{1}{2} \cdot \alpha \cdot h = \frac{1}{2} \cdot \alpha \cdot \frac{\beta}{2} = \frac{\alpha\beta}{4}

Using two sides and the included angle, the same area is

Area=12⋅r1⋅r2⋅sin⁡θ=12⋅5⋅4⋅sin⁡θ=10sin⁡θ\text{Area} = \frac{1}{2} \cdot r_1 \cdot r_2 \cdot \sin\theta = \frac{1}{2} \cdot 5 \cdot 4 \cdot \sin\theta = 10\sin\theta

Equating the two expressions,

αβ4=10sin⁡θ\frac{\alpha\beta}{4} = 10\sin\theta

Hence,

αβ=40sin⁡θ\alpha\beta = 40\sin\theta

Now substitute the given value:

αβ=40(638)=563\alpha\beta = 40\left(\frac{\sqrt{63}}{8}\right) = 5\sqrt{63}

Therefore,

(αβ)2=(563)2=25×63=1575(\alpha\beta)^2 = (5\sqrt{63})^2 = 25 \times 63 = 1575

Therefore, the value of (αβ)2(\alpha\beta)^2 is 15751575.

Common mistakes

  • Confusing the angle given in the question with the angle at the centers. The stated angle is between the two radii drawn from an intersection point, so it is ∠O1PO2\angle O_1PO_2, not an angle at O1O_1 or O2O_2. Use the area formula with sides O1PO_1P and O2PO_2P.

  • Taking the common chord length as the altitude itself. If the altitude from the intersection point to the line of centers is hh, then the chord length is β=2h\beta = 2h. Missing this factor of 22 gives the wrong relation for the area.

  • Using the detailed chord-length formula in terms of α\alpha unnecessarily. The product αβ\alpha\beta follows directly from the area of triangle O1PO2O_1PO_2, so introducing extra algebra can make the solution longer and more error-prone.

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