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MCQEasyJEE 2024Circle Equation & Properties

JEE Mathematics 2024 Question with Solution

Consider a circle (x−α)2+(y−β)2=50(x - \alpha)^2 + (y - \beta)^2 = 50, where α,β>0\alpha, \beta > 0. If the circle touches the line y+x=0y + x = 0 at point PP, whose distance from the origin is 424\sqrt{2}, then (α+β)2(\alpha + \beta)^2 is equal to:

  • A

    100100

  • B

    150150

  • C

    5050

  • D

    200200

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The circle is (x−α)2+(y−β)2=50(x - \alpha)^2 + (y - \beta)^2 = 50, so its center is C(α,β)C(\alpha, \beta) and radius is 50=52\sqrt{50} = 5\sqrt{2}.

Find: The value of (α+β)2(\alpha + \beta)^2.

Since the circle touches the line y+x=0y + x = 0, the perpendicular distance from the center C(α,β)C(\alpha, \beta) to this line equals the radius.

Using the distance formula from point (α,β)(\alpha, \beta) to the line x+y=0x + y = 0:

∣α+β∣12+12=∣α+β∣2\frac{|\alpha + \beta|}{\sqrt{1^2 + 1^2}} = \frac{|\alpha + \beta|}{\sqrt{2}}

Equate this distance to the radius:

∣α+β∣2=52\frac{|\alpha + \beta|}{\sqrt{2}} = 5\sqrt{2}

So,

∣α+β∣=52⋅2=10|\alpha + \beta| = 5\sqrt{2} \cdot \sqrt{2} = 10

Since α,β>0\alpha, \beta > 0, we have α+β>0\alpha + \beta > 0. Therefore,

α+β=10\alpha + \beta = 10

Hence,

(α+β)2=102=100(\alpha + \beta)^2 = 10^2 = 100

Therefore, the correct option is A.

Using the touching point information

Given: The point of contact PP lies on the line x+y=0x + y = 0 and its distance from the origin is 424\sqrt{2}.

Find: The value of (α+β)2(\alpha + \beta)^2.

A point on the line x+y=0x + y = 0 can be written as P(a,−a)P(a,-a).

Using the distance from the origin:

a2+(−a)2=42\sqrt{a^2 + (-a)^2} = 4\sqrt{2} 2a2=42\sqrt{2a^2} = 4\sqrt{2} ∣a∣=4|a| = 4

So the possible touching points are (4,−4)(4,-4) and (−4,4)(-4,4).

This confirms that the touching point lies on the given line, but to find α+β\alpha + \beta we use the tangency condition: distance from the center to the line equals the radius.

The radius is

r=50=52r = \sqrt{50} = 5\sqrt{2}

And the distance from C(α,β)C(\alpha,\beta) to x+y=0x+y=0 is

∣α+β∣2\frac{|\alpha+\beta|}{\sqrt{2}}

Therefore,

∣α+β∣2=52\frac{|\alpha+\beta|}{\sqrt{2}} = 5\sqrt{2} ∣α+β∣=10|\alpha+\beta| = 10

Since α,β>0\alpha,\beta>0,

α+β=10\alpha+\beta=10

Thus,

(α+β)2=100(\alpha+\beta)^2 = 100

Therefore, the answer is 100100, so the correct option is A.

Common mistakes

  • Using the point of contact coordinates to form unnecessary equations for α\alpha and β\beta. The direct tangency condition is enough: distance from the center to the line must equal the radius.

  • Forgetting the modulus in the point-to-line distance formula. The correct expression is ∣α+β∣2\frac{|\alpha+\beta|}{\sqrt{2}}, not α+β2\frac{\alpha+\beta}{\sqrt{2}} initially.

  • Not using the condition α,β>0\alpha, \beta > 0. After obtaining ∣α+β∣=10|\alpha+\beta|=10, this condition implies α+β=10\alpha+\beta=10, not −10-10.

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