Question

given a circle with the equation x² + y² - 8x + 12y + 16 = 0, what is the coordinate of its center and the length of its radius? enter your answers in the boxes. the coordinate of the circles center is ( , ). the circles radius is units.

Explanation:

Step1: Rewrite the circle - equation in standard form

The general equation of a circle is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and its standard - form is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center of the circle and \(r\) is the radius. Given \(x^{2}+y^{2}-8x + 12y+16 = 0\). Complete the square for \(x\) and \(y\) terms:
\[

$$\begin{align*} x^{2}-8x+y^{2}+12y&=-16\\ x^{2}-8x + 16+y^{2}+12y+36&=-16 + 16+36\\ (x - 4)^{2}+(y + 6)^{2}&=36 \end{align*}$$

\]

Step2: Identify the center and radius

Comparing \((x - 4)^{2}+(y + 6)^{2}=36\) with \((x - a)^{2}+(y - b)^{2}=r^{2}\), we have \(a = 4\), \(b=-6\), and \(r^{2}=36\). So the center of the circle is \((4,-6)\) and the radius \(r = 6\).

Answer:

The coordinate of the circle's center is \((4,-6)\)
The circle's radius is \(6\) units