Standard equation of circle with center of (6,8) that passes through point (-1,4)

Question

Standard equation of circle with center of (6,8) that passes through point (-1,4)

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Peyton 7 hours 2021-09-12T03:20:13+00:00 1 Answer 0

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    2021-09-12T03:21:58+00:00

    Answer:

    The result should be:

    (x-6)^2+(y-8)^2=\sqrt{65}

    Step-by-step explanation:

    -The equation of a circle:

    (x-h)^2+(y-k)^2=r^2 where the center is (h,k), the point (x,y) and the radius known as r.

    -Use the center (6,8) and the point (-1,4) for this equation:

    (-1-6)^2+(4-8)^2=r^2

    -Then, you solve for the radius and get the equation:

    (-1-6)^2+(4-8)^2=r^2

    (-7)^2+(-4)^2 =r^2

    49 + 16 = r^2

    65 = r^2

    \sqrt{65} =\sqrt{r^2}

    \sqrt{65} = r

    -Result:

    (x-6)^2+(y-8)^2=\sqrt{65}

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45:7+7-4:2-5:5*4+35:2 =? ( )