How do you find the center and radius of the circle given #x^2-12x+84=-y^2+16y#?

2 Answers
Oct 13, 2016

Answer:

Center (6, 8) and radius = 4

Explanation:

rewrite equation as
x^2 - 12x +y^2-16y +84 =0
(x^2 -12x +36) + (y^2-16y +64) +84 -36-64 =0
(x-6)^2 +(y-8)^2 -16 =0
(x-6)^2 +(y-8)^2 = 16 = 4^2
So center is (6,8) and radius 4

Oct 13, 2016

Answer:

Write the equation as: #(x - 6)^2 + (y - 8)^2 = 4^2#
Center: #(6,8)#
Radius: 4

Explanation:

You need to put the equation into standard form:

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

because we know that #(h, k)# is the center and #r# is the radius.

In the given equation, move the constant term to the right and all of the other terms to the left:

#x^2 - 12x + y^2 - 16y = -84#

Add #h^2 and k^2# to both sides:

#x^2 - 12x + h^2 + y^2 - 16y + k^2= -84 +h^2 + k^2#

Because #(x - h)^2 = x^2 - 2hx + h^2#, we can set middle term of this pattern equal to the middle term of the given equation and then find the value of #h# and #h^2#:

#-2hx = -12x#

#h = 6, h^2 = 36#

Substitute 36 for every #h^2#

#x^2 - 12x + 36 + y^2 - 16y + k^2= -84 +36 + k^2#

We know that the left side is a perfect square and the right side simplifies a bit:

#(x - 6)^2 + y^2 - 16y + k^2= -48 + k^2#

We do the same thing for the middle term to find #k# and #k^2#:

#-2ky = -16y#

#k = 8, k^2 = 64#

Substitute 64 for every #k^2#:

#(x - 6)^2 + y^2 - 16y + 64= -48 + 64#

We know that the left is a perfect square and the right side simplifies:

#(x - 6)^2 + (y - 8)^2 = 16#

Make the radius obvious by writing the ride side as a square:

#(x - 6)^2 + (y - 8)^2 = 4^2#