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If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle?
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If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle?

User KSTN
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The midpoint of a diameter is the centre of the circle.

The centre is important for finding the equation.

Midpoint = (-10, 1)

The radius is also important

I will include it in the image

Equation of circle (x +10)^2 + (y -2)^2 = 5
If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the-example-1
User Sarreph
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A(-8,0) B(-12,12)the center is w((-8-12)/2 , (0+12)/2)....(midel [ AB]w(-10 ,6)the ridus is : r = AB/2AB = √(-12+8)²+(12-0)² = √16+144 =√160/2an equation in standard form equation of the circle :(x+10)²+(y-6)² = (√160/2)²=160/4 = 40

User Ibn Rushd
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6.1k points
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