Question:
What is the equation of the circle (h, k) = (3, 6) , radius 4?
Ax² + y² - 12xy + 29 = 0
Bx² + y² - 6x - 12y + 29 = 0
Cx² + y² - 6xy + 29 = 0
Dx² + y² - + 29 = 0
Answer:
B. x² + y² - 6x - 12y + 29 = 0
Explanation:
If the circle with base (h, k) passes through the point (x, y) then equation of the circle = (x - h)² + (y - k)² = r² (x - 3)² + (y - 6)² = 4² x² - 6x + 3² + y² - 12y + 6² =16 x² + y² - 6x - 12y + 29 = 0