The locus of the mid-point of a chord of the circle which subtends a right angle at the origin, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
If (h,k) is the midpoint of a chord of x²+y²=4, the equation of that chord is hx+ky=h²+k² (T=S₁). For the chord to subtend a right angle at the origin, homogenize the circle with the chord: x²+y²=4 combined with (hx+ky)/(h²+k²)=1 gives x²+y²−4·(hx+ky)²/(h²+k²)²·(h²+k²)=0, i.e., x²+y²−4(hx+ky)²/(h²+k²)=0. The angle at origin is 90° when the sum of coefficients of x² and y² in this homogenized equation equals zero: (1−4h²/(h²+k²))+(1−4k²/(h²+k²))=0 → 2−4(h²+k²)/(h²+k²)=0 → 2−4=−2≠0... Let me redo: coefficients are (1−4h²/(h²+k²)) and (1−4k²/(h²+k²)), their sum = 2−4(h²+k²)/(h²+k²)=2−4=−2, so each must satisfy the 90° condition via coeff sum=0 giving 2(h²+k²)=4, i.e., h²+k²=2. The locus is x²+y²=2.
[1984]