Question

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

$x^{2}+y^{2}+14x - 8y + 16 = 0$

answer attempt 1 out of 10

center: ( , )

radius:

Explanation:

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

$$x^{2}+14x + y^{2}-8y+16 = 0$$

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

For \(x^{2}+14x\), we use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here \(a=x\) and \(2b = 14\Rightarrow b = 7\). So \(x^{2}+14x=(x + 7)^{2}-49\)

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

For \(y^{2}-8y\), using \((a - b)^2=a^{2}-2ab + b^{2}\). Here \(a = y\) and \(2b=8\Rightarrow b = 4\). So \(y^{2}-8y=(y - 4)^{2}-16\)

Step4: Substitute back into the equation

$$(x + 7)^{2}-49+(y - 4)^{2}-16+16=0$$
$$(x + 7)^{2}+(y - 4)^{2}=49$$

Answer:

Center: \((-7,4)\)
Radius: \(7\)