Find the equation of a circle with a center at (7, 2) and a point on the circle at (-2, -5).
1 answer:
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Answer:
a (3,-1)
Step-by-step explanation:
The number that "completes the square" is the square of half the x-coefficient, (-6/2)^2 = 9. Rearranging the given function to include the square trinomial, we have ...
f(x) = x^2 -6x +9 -1 . . . . . . . here, we have 8 = 9 - 1
f(x) = (x -3)^2 -1 . . . . . . . . . . vertex form
Comparing this to the generic vertex form ...
f(x) = (x -h)^2 +k . . . . . . . vertex at (h, k)
we see that h=3 and k=-1.
The vertex is (h, k) = (3, -1).
