How do you write an equation for a circle given center (8,-9) and passes through (21,22)?

1 Answer
Douglas K.
Dec 12, 2016

Please see the explanation for steps leading to the equation.

Explanation:

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

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

We are given that the center point is #(8, -9)#, therefore, we substitute 8 for h and -9 for k into equation [1]:

#(x - 8)^2 + (y - -9)^2 = r^2" [2]"#

NOTE: One can write #(y - -9)^2# as #(y + 9)^2# but doing so can cause errors when attempting to obtain the center point. Therefore, I do not recommend it.

To find the value of r, substitute the given point, 21 for x and 22 for y, into equation [2]:

#(21 - 8)^2 + (22 - -9)^2 = r^2#

#(13)^2 + (31)^2 = r^2#

#169 + 961 = r^2#

#r^2 = 1130#

#r = sqrt(1130)#

Substitute #sqrt(1130)# for r in equation [2]:

#(x - 8)^2 + (y - -9)^2 = (sqrt(1130))^2" [3]"#

Equation [3] is the standard form for the circle.

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