How do you write the equation for a circle with center C(7,-4) and passes through P(-3,3)?

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Answer 1 · 2016-11-18T00:19:41

Please see the explanation.

The standard equation for a circle is:

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

where #(x, y)# is any point on the circle, #(h, k)# is the center, and r it the radius.

GIven the center is #(7, -4)#, substitute for h and k in the general form:

#(x - 7)^2 + (y - -4)^2 = r^2#

Note: I do not recommend that you change the second term to be #(y + 4)^2#; doing so can cause confusion, when asked for the center.

To find the value of r substitute the given point:

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

#r^2 = 149#

#r = sqrt(149)#

The equation is:

#(x - 7)^2 + (y - -4)^2 = (sqrt(149))^2#