SOLUTION: Find the standard form of the equation for the circle with the following properties. Endpoints of a diameter are (8,4) and (12,-4)

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Question 1066044: Find the standard form of the equation for the circle with the following properties. Endpoints of a diameter are (8,4) and (12,-4)
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Center is the midpoint of the diameter.
x=%288%2B12%29%2F2=10
y=%284%2B%28-4%29%29%2F2=0

Radius SQUARED is cross%28%288-12%29%5E2%2B%284-%28-4%29%29%5E2=16%2B64=80%29 %281%2F2%29%5E2%2Asqrt%28%2812-8%29%5E2%2B%28-4-4%29%5E2%29%5E2
%281%2F4%29%28sqrt%2816%2B64%29%29%5E2
%281%2F4%2980
20


EQUATION: %28x-10%29%5E2%2By%5E2=20

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Find the standard form of the equation for the circle with the following properties. Endpoints of a diameter are (8,4) and (12,-4)
Coordinates of midpoint of diameter, or center of circle: (10, 0)

Equation of circle: %28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2

%288+-+10%29%5E2+%2B+%284+-+0%29%5E2+=+r%5E2 ------- Substituting (8, 4) for (x, y), and (10, 0) for (h, k)

%28-+2%29%5E2+%2B+4%5E2+=+r%5E2

matrix%281%2C5%2C+4+%2B+16+=+r%5E2%2C+%22======%3E%22%2C+r%5E2%2C+%22=%22%2C+20%29

Equation of this circle: highlight_green%28%28x+-+10%29%5E2+%2B+y%5E2+=+20%29