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MCQMediumJEE 2023Circle Equation & Properties

JEE Mathematics 2023 Question with Solution

Let the centre of a circle C be (α\alpha, β\beta) and its radius r<8r < 8. Let 3x+4y−243x + 4y - 24 and 3x−4y−323x - 4y - 32 be two tangents and 4x+3y=14x + 3y = 1 be a normal to C. Then the value of (α−β\alpha - \beta) is equal to:

  • A

    55

  • B

    66

  • C

    77

  • D

    99

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The centre of the circle is (α\alpha, β\beta), the radius satisfies r<8r < 8, the tangents are 3x+4y−24=03x + 4y - 24 = 0 and 3x−4y−32=03x - 4y - 32 = 0, and the line 4x+3y=14x + 3y = 1 is a normal to the circle.

Find: The value of α−β\alpha - \beta.

Since 4x+3y=14x + 3y = 1 is a normal to the circle, the centre lies on it. Hence,

4α+3β=14\alpha + 3\beta = 1

Also, the perpendicular distances of the centre from the two tangents are equal. So,

∣3α+4β−24∣32+42=∣3α−4β−32∣32+(−4)2\frac{|3\alpha + 4\beta - 24|}{\sqrt{3^2 + 4^2}} = \frac{|3\alpha - 4\beta - 32|}{\sqrt{3^2 + (-4)^2}}

which gives

∣3α+4β−24∣=∣3α−4β−32∣|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32|

Taking the valid case,

3α+4β−24=−(3α−4β−32)3\alpha + 4\beta - 24 = -(3\alpha - 4\beta - 32) 3α+4β−24=−3α+4β+323\alpha + 4\beta - 24 = -3\alpha + 4\beta + 32 6α=566\alpha = 56 α=283\alpha = \frac{28}{3}

Using 4α+3β=14\alpha + 3\beta = 1,

4(283)+3β=14\left(\frac{28}{3}\right) + 3\beta = 1 1123+3β=1\frac{112}{3} + 3\beta = 1 9β=−1099\beta = -109 β=−1099\beta = -\frac{109}{9}

This leads to a radius greater than 88, so this case is rejected as indicated in the solution.

Now take the other valid equality case and solve with the normal equation. From the solution, this gives

α=1,β=−1\alpha = 1, \quad \beta = -1

Therefore,

α−β=1−(−1)=2\alpha - \beta = 1 - (-1) = 2

However, the solution concludes with "Therefore, α−β+r=7\alpha - \beta + r = 7" and marks the correct option as C. Since the source solution is internally inconsistent but explicitly declares C as correct, the recorded answer is C.

Therefore, the correct option is C.

Consistency Note

The solution contains contradictory working. It first obtains one rejected case, then states α=1\alpha = 1 and β=−1\beta = -1, and finally concludes α−β+r=7\alpha - \beta + r = 7, which does not match the asked quantity α−β\alpha - \beta. The solution explicitly says The Correct Option is C, so the answer is taken as C according to the source authority, while noting the inconsistency in the working.

Common mistakes

  • Equating the two tangent distances without using absolute value. Distances from a point to two lines are non-negative, so signs matter. Use ∣3α+4β−24∣=∣3α−4β−32∣|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32| before splitting into cases.

  • Forgetting that the centre lies on the normal. Since 4x+3y=14x + 3y = 1 is a normal to the circle, the centre must satisfy this line equation. Substitute the centre coordinates into the line first.

  • Ignoring the condition r<8r < 8. One algebraic case may satisfy the equal-distance condition but produce a radius larger than 88. That branch must be rejected.

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