If the circles and , , intersect at the points and , then the line passes through and , for
- A.no values of
- B.exactly one value of
- C.exactly two values of
- D.infinitely many values of
Show answer and solution
Correct option: A
Solution
The radical axis (common chord PQ) is found by subtracting the two circle equations. Rewrite the second circle as x² + y² + Kx + (3/2)y - 1/2 = 0. Subtract from the first: (5K - K)x + (2 - 3/2)y + (K + 1/2) = 0, giving 4Kx + (1/2)y + (K + 1/2) = 0, or equivalently 8Kx + y + 2K + 1 = 0.
For the line 4x + 5y - K = 0 to be the same line PQ, it must be proportional to 8Kx + y + (2K+1) = 0. Comparing coefficients: 8K/4 = 1/5 = (2K+1)/(-K), so from the first two ratios 2K = 1/5, giving K = 1/10, but from the last two ratios 1/5 = (2K+1)/(-K) → -K = 10K+5 → K = -5/11. These are contradictory, so no single value of K satisfies all three ratios simultaneously, meaning the line never coincides with PQ for any K ∈ R.
For the line 4x + 5y - K = 0 to be the same line PQ, it must be proportional to 8Kx + y + (2K+1) = 0. Comparing coefficients: 8K/4 = 1/5 = (2K+1)/(-K), so from the first two ratios 2K = 1/5, giving K = 1/10, but from the last two ratios 1/5 = (2K+1)/(-K) → -K = 10K+5 → K = -5/11. These are contradictory, so no single value of K satisfies all three ratios simultaneously, meaning the line never coincides with PQ for any K ∈ R.
[JEE Main 2019 · 10 April · Shift I]