What is the equation of a circle in general form if the center is 2,6 passes through the point 3,1? Thanks

1 Answer
Teddy
Jun 20, 2018

(x-2)^2+(y-6)^2=26

Explanation:

The general form of the equation of a circle is (x-h)^2+(y-k)^2=r^2 where (h,k) are the coordinates of the center and r is the length of the radius.

We are given that the center is at (2,6) so we need to figure out the length of the radius. Since we know the circle passes through the point (3,1), we know the radius is just the distance between (2,6) and (3,1). In the equation, we end up using r^2 instead of r which makes our calculations a bit easier:

"Distance"=r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
r^2=(x_2-x_1)^2+(y_2-y_1)^2
r^2=(2-3)^2+(6-1)^2
r^2=26

Now that we have everything, we just substitute our values into the equation of a circle: (x-2)^2+(y-6)^2=26

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