If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle? A) (x − 10)2 + (y +

Question

If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle? A) (x − 10)2 + (y + 1)2 = 5 B) (x + 10)2 + (y − 1)2 = 5 C) (x − 10)2 + (y + 1)2 = 5 D) (x + 10)2 + (y − 1)2 = 5

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Faith 2 hours 2021-09-11T04:59:29+00:00 1 Answer 0

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    2021-09-11T05:00:53+00:00

    Answer:

    Step-by-step explanation:

    We are given the diameter of the circle.  By definition, the diameter goes through the circle right at its center.  We can use the midpoint formula to find the center of the diameter which will give us the center of our circle.

    M=(\frac{-8+(-12)}{2}, \frac{0+2}{2}) which simplifies to

    M=(\frac{-20}{2},\frac{2}{2})  which gives us the midpoint

    M = (-10, 1)  That’s the h and k of our center.  We will use that along with one of the given points to solve for r.  Plugging into the standard form of a circle (which I’m assuming you know):

    (-8+10)^2+(0-1)^2=r^2 and

    (2)^2+(-1)^2=r^2  so

    r^2=5

    Now we’ll use that with the center we found and write the equation:

    (x+10)^2+(y-1)^2=5

    That’s choice B

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