the midpoint of the diameter is the center of the circle: (6,4)
The radius is half the diameter: 1
(x-6)^2 + (y-4)^2 = 1
Find the equation of a circle which diameter has the end point A (5, 4), B (7,4 ).
8 answers
Correct steve
No the answer is
X^2 + y^2 - r^2 - 12x - 8y + 52 = 0
X^2 + y^2 - r^2 - 12x - 8y + 52 = 0
The answer is
X^2 + y^2 - r^2 - 12x - 8y + 52 = 0
X^2 + y^2 - r^2 - 12x - 8y + 52 = 0
My money is on Steve's correct answer.
Steve got it but it did nt reach quarter of d workings
X²+Y² -12x -8y -53 =0
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