Find the complete set of values of p for which the equation x2 − 2px + y2 − 6y − p2 + 8p + 9 = 0 describes a circle in the xy-plane.
- Ap < −94
- B0 < p < 4
- C−1 < p < 9
- Dp < 0 or p > 4
- Ep < −1 or p > 9
- Fall real values of p
Show the answer and worked solution
answer · D
- Ap < −94
- B0 < p < 4
- C−1 < p < 9
- Dp < 0 or p > 4
- Ep < −1 or p > 9
- Fall real values of p
Completing both squares gives (x−p)2 − p2 + (y−3)2 − 9 − p2 + 8p + 9 = 0, that is (x−p)2 + (y−3)2 = 2p2 − 8p. This is a genuine circle only when the right-hand side is strictly positive: 2p(p−4) > 0, so p < 0 or p > 4.