Question

find the center and radius of the circle represented by the equation below.

$x^{2}+y^{2}-8x - 12y + 27 = 0$

answer attempt 1 out of 10

center: ( , )

radius:

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Explanation:

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

$$(x^{2}-8x)+(y^{2}-12y)+27 = 0$$

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

For \(x^{2}-8x\), we have \((x - 4)^{2}=x^{2}-8x + 16\).

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

For \(y^{2}-12y\), we have \((y - 6)^{2}=y^{2}-12y+36\).

Step4: Rewrite the equation

$$(x - 4)^{2}-16+(y - 6)^{2}-36+27 = 0$$
$$(x - 4)^{2}+(y - 6)^{2}-25 = 0$$
$$(x - 4)^{2}+(y - 6)^{2}=25$$
The standard form of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.

Answer:

Center: \((4,6)\)
Radius: \(5\)