If a tangent to the circle intersects the coordinate axes at distinct points and , then the locus of the mid-point of is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The tangent to x² + y² = 1 in intercept form is x/a + y/b = 1, with the condition 1/a² + 1/b² = 1.
P = (a, 0) and Q = (0, b), so the midpoint is (h, k) = (a/2, b/2), giving a = 2h, b = 2k.
Substituting into the condition: 1/(2h)² + 1/(2k)² = 1, i.e., 1/(4h²) + 1/(4k²) = 1, so h² + k² = 4h²k².
The locus is x² + y² - 4x²y² = 0.
P = (a, 0) and Q = (0, b), so the midpoint is (h, k) = (a/2, b/2), giving a = 2h, b = 2k.
Substituting into the condition: 1/(2h)² + 1/(2k)² = 1, i.e., 1/(4h²) + 1/(4k²) = 1, so h² + k² = 4h²k².
The locus is x² + y² - 4x²y² = 0.
[JEE Main 2019 · 9 April · Shift I]