The standard form of a circle with centre (3, -2) and radius 5 is:
A$(x-3)^2 + (y+2)^2 = 25$
B$(x+3)^2 + (y-2)^2 = 25$
C$(x-3)^2 + (y-2)^2 = 5$
D$x^2 + y^2 + 6x - 4y - 4 = 0$
Answer & Solution
Correct answer: A. $(x-3)^2 + (y+2)^2 = 25$
Standard form: (x − h)² + (y − k)² = r². With (h, k) = (3, -2), r = 5: (x − 3)² + (y − (-2))² = 5² → (x − 3)² + (y + 2)² = 25.
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