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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