Question

find the center and radius of the circle represented by the equation below.
$x^{2}+y^{2}-10x - 18y+57 = 0$
answer attempt 1 out of 2
center: ( )
radius:

Explanation:

Step1: Group \(x\) and \(y\) terms

$$x^{2}-10x + y^{2}-18y=-57$$

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

For \(x^{2}-10x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2b = 10\) (\(b = 5\)), so \(x^{2}-10x=(x - 5)^{2}-25\)

Step3: Complete the square for \(y\) terms

For \(y^{2}-18y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2b = 18\) (\(b = 9\)), so \(y^{2}-18y=(y - 9)^{2}-81\)

Step4: Substitute back into the equation

$$(x - 5)^{2}-25+(y - 9)^{2}-81=-57$$
$$(x - 5)^{2}+(y - 9)^{2}-106=-57$$
$$(x - 5)^{2}+(y - 9)^{2}=49$$

Answer:

Center: \((5,9)\)
Radius: \(7\)