How do you find the equation for the circle centered at (4, 4); passing through (7, 14)?

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Answer 1 · 2016-07-31T08:43:20

#(x-4)^2+(y-4)^2 = 109#

or

#x^2+y^2-8x-8y-77 = 0#

The general equation of a circle with centre #(h, k)# and radius #r# may be written:

#(x-h)^2+(y-k)^2 = r^2#

In our example, we know #(h, k) = (4, 4)#, so we just need to determine #r^2#.

#(x-4)^2+(y-4)^2 = r^2#

Since the circle passes through #(7, 14)#, that pair of values for #x# and #y# must satisfy the equation and we have:

#r^2 = (7-4)^2+(14-4)^2 = 3^2+10^2 = 109#

So the equation of the circle may be written:

#(x-4)^2+(y-4)^2 = 109#

We can reexpress this in standard polynomial form by expanding and rearranging:

#color(blue)(x^2-8x+16)+color(green)(y^2-8y+16) = 109#

So:

#x^2+y^2-8x-8y-77 = 0#