Question Detail
Question
Find the equation of the circle whose center is the same as that of the circle x² + y² - 6x + 2y + 4 = 0 and which passes through the point (7,4).
Correct Answer
Option B is the correct answer.
Detailed Explanation
First find center of given circle: x² - 6x + y² + 2y + 4 = 0 → (x² - 6x + 9) + (y² + 2y + 1) = -4 + 9 + 1 = 6 → (x-3)² + (y+1)² = 6. Center = (3,-1). New circle has same center and passes through (7,4). Radius² = (7-3)² + (4+1)² = 16 + 25 = 41. Equation: (x-3)² + (y+1)² = 41 → x² - 6x + 9 + y² + 2y + 1 = 41 → x² + y² - 6x + 2y + 10 - 41 = 0 → x² + y² - 6x + 2y - 31 = 0.
Hint
Complete the square to find center, then use distance formula.