Question

Consider the following equation of a circle find the center, (h,k)

Answer 1

Given the equation of a circle:

$x^2+y^2-4x+4y+4=0$

Let's find the center of the circle, (h, k).

To find the center of the circle, rewrite the equation in standard form:

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

Where:

(h, k) is the center.

Rewrite the equation:

$(x^2-4x)+y^2+4y+4=0$

Complete the square for the two groups:

x²-4x and y²+ 4y.

Apply the formula:

$a(x+d)^2+e$

Now, we have the following:

$\begin{gathered} x^2-4x \\ Where: \\ a=1 \\ b=-4 \\ c=0 \\ \\ \text{ To solve for d, we have:} \\ d=\frac{b}{2a}=\frac{-4}{2(1)}=-2 \\ \\ \text{ To solve for e, we have:} \\ e=c-\frac{b^2}{4a}=0-\frac{-4^2}{4(1)}=-\frac{16}{4}=-4 \\ \\ \text{ FOr the first group, we have:} \\ (x-2)^2-4 \end{gathered}$

Complete the square for the second group:

$\begin{gathered} y^2+4y \\ \text{ Where:} \\ a=1 \\ b=4 \\ c=0 \\ \\ \text{ We have:} \\ d=\frac{b}{2a}=\frac{4}{2(1)}=2 \\ \\ e=c-\frac{b^2}{4a}=0-\frac{4^2}{4(1)}=0-\frac{16}{4}=-4 \\ \text{ For the second group, we have:} \\ (y+2)^2-4 \\ \end{gathered}$

Now, combine the expressions in the original equation

$\begin{gathered} ((x-2)^2-4)+((y+2)^2-4)+4=0 \\ \\ (x-2)^2-4+(y+2)^2-4+4=0 \end{gathered}$

Combine like terms and move the constants to the right side of the equation:

$\begin{gathered} (x-2)^2+(y+2)^2-4+4-4=0 \\ \\ (x-2)^2+(y+2)^2-4=0 \\ \\ (x-2)^2+(y+2)^2=4 \end{gathered}$

Therefore, the equation of the circle in standard form is:

$(x-2)^{2}+(y+2)^{2}=4$

Where:

h = 2

k = -2

Therefore, the center of the circle is:

(h, k) ==> (2, -2)

ANSWER:

(2, -2)