How do you find the center and radius of the circle #x^2-8x+y^2-12y= -51#?

Question

Lincoln Anderson · Accepted Answer

The center is the point #(4, 6)# and the radius, #r = 1# Add #k^2# and #h^2# to both sides: #x^2 - 8x + h^2 + y^2 - 12y + k^2 = h^2 + k^2 - 51# Use the pattern #(x - h)^2 = x^2 - 2hx + h^2# to find the value of #h# and #h^2#: #x^2 - 2hx + h^2 = x^2 - 8x + h^2# #-2hx = -8x# #h = 4# #h^2 = 16# Write the left side as a perfect square and substitute 16 for #h^2# on the right: #(x - 4)^2 + y^2 - 12y + k^2 = 16 + k^2 - 51# Use the pattern #(y - k)^2 = y^2 - 2ky + k^2# to find the value of #k# and #k^2#: #y^2 - 2ky + k^2 = y^2 - 12y + k^2# #-2ky=-12y# #k = 6# #k^2 = 36# Write the left side as a perfect square and substitute 36 for #k^2# on the right: #(x - 4)^2 + (y - 6)^2 = 16 + 36 - 51# #(x - 4)^2 + (y - 6)^2 = 1#

Elias Ackerman · Answer

undefined To find the center and radius of the circle represented by the equation $x^2 - 8x + y^2 - 12y = -51$, we complete the square for both the $x$ and $y$ terms, and then rearrange the equation into standard form for a circle, which is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ represents the center of the circle and $r$ represents the radius.

First, complete the square for the $x$ and $y$ terms:

For $x^2 - 8x$:
1. $x^2 - 8x$
2. Half of the coefficient of $x$ is $-4$, so add $(-4)^2 = 16$ inside the parentheses:
   $$x^2 - 8x + 16$$

For $y^2 - 12y$:
1. $y^2 - 12y$
2. Half of the coefficient of $y$ is $-6$, so add $(-6)^2 = 36$ inside the parentheses:
   $$y^2 - 12y + 36$$

Now, rewrite the equation with the completed square terms:
$$x^2 - 8x + 16 + y^2 - 12y + 36 = -51 + 16 + 36$$
$$(x - 4)^2 + (y - 6)^2 = 1$$

Comparing this equation to the standard form of a circle, we see that the center $(h, k)$ is $(4, 6)$, and the radius $r$ is $\sqrt{1}$, which simplifies to $1$. Therefore, the center of the circle is $(4, 6)$, and its radius is $1$.