Question

$x^{2}+y^{2}-4x + 4y+7 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$ for:
$h=$
and
$k=$
and
$r=$
graph the circle.

Explanation:

Step1: Rewrite the equation by completing the square for x and y terms.

Given the equation \(x^{2}+y^{2}-4x + 4y+7 = 0\).
For the x - terms: \(x^{2}-4x=(x - 2)^{2}-4\) (using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2ab = 4x\), so \(b = 2\)).
For the y - terms: \(y^{2}+4y=(y + 2)^{2}-4\) (using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\), \(2ab = 4y\), so \(b = 2\)).

Substitute these into the original equation:
\((x - 2)^{2}-4+(y + 2)^{2}-4 + 7=0\)

Step2: Simplify the equation to the standard form of a circle.

Simplify the left - hand side:
\((x - 2)^{2}+(y + 2)^{2}-4-4 + 7=0\)
\((x - 2)^{2}+(y + 2)^{2}-1 = 0\)
\((x - 2)^{2}+(y + 2)^{2}=1\)

The standard form of the equation of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Comparing \((x - 2)^{2}+(y + 2)^{2}=1\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\):

• For the x - coordinate of the center: \(h = 2\) (since \(x-2=x - h\) implies \(h = 2\)).
• For the y - coordinate of the center: \(y+2=y - k\) implies \(k=- 2\) (because \(y+2=y-(-2)\)).
• For the radius: \(r^{2}=1\), so \(r = 1\) (since \(r>0\)).

Answer:

\(h = 2\), \(k=-2\), \(r = 1\)