Question

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$x^{2}+y^{2}-4x + 8y+19 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$ for:
$h = 2$
and
$k=-4$
and
$r = 1$
graph the circle.
question help: video

Explanation:

Step1: Complete the square for \(x\) terms

Group \(x\) terms: \(x^{2}-4x=(x - 2)^{2}-4\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 2\)).

Step2: Complete the square for \(y\) terms

Group \(y\) terms: \(y^{2}+8y=(y + 4)^{2}-16\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 4\)).

Step3: Rewrite the circle equation

Substitute into \(x^{2}+y^{2}-4x + 8y+19 = 0\):
\((x - 2)^{2}-4+(y + 4)^{2}-16+19 = 0\)
\((x - 2)^{2}+(y + 4)^{2}-1=0\)
\((x - 2)^{2}+(y + 4)^{2}=1\)

Answer:

The center \((h,k)=(2,-4)\) and radius \(r = 1\).