How do you find the center and radius for # x^2 + y^2 +14x = 0#?

1 Answer
Oct 25, 2016

Answer:

Please see the explanation.

Explanation:

The standard form for the equation of a circle is:

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

Substitute #(y - 0)^2# for #y^2#

#x^2 + 14x + (y - 0)^2 = 0#

Using the pattern #(x - h)^2 = x^2 - 2hx + h^2#, we observe that we must begin the process of completing the square by adding #h^2# to both sides:

#x^2 + 14x +h^2 + (y - 0)^2 = h^2#

For this circle, the #h^2# term is, also the #r^2# term, because the we will soon find that the center is offset to the left the same distance as the radius.

We can find the value #h# by setting the right side of the pattern equal to the first 3 terms in the equation:

#x^2 - 2hx + h^2 = x^2 + 14x +h^2#

The square terms cancel:

#-2hx = 14x#

#h = -7 and h^2 = 49

Substitute #(x - -7)^2# for the corresponding terms on the left and

#(x - -7)^2 + (y - 0)^2 = 49#

The radius should be represented as a positive number squared.

#(x - -7)^2 + (y - 0)^2 = 7^2#

The center is found by observation #(-7, 2)#