Question

question 9 of 39
what are the center and radius of the circle defined by the equation
$x^{2}+y^{2}-6x + 8y+21 = 0$?
a. center $(3,-4)$; radius 2
b. center $(-3,4)$; radius 4
c. center $(-3,4)$; radius 2
d. center $(3,-4)$; radius 4

Explanation:

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

Given \(x^{2}-6x\), we have \((x - 3)^{2}=x^{2}-6x + 9\).

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

Given \(y^{2}+8y\), we have \((y + 4)^{2}=y^{2}+8y+16\).

Step3: Rewrite the circle equation

The original equation \(x^{2}+y^{2}-6x + 8y+21 = 0\) can be rewritten as \((x - 3)^{2}-9+(y + 4)^{2}-16 + 21=0\).
Simplify it: \((x - 3)^{2}+(y + 4)^{2}-25 + 21=0\), then \((x - 3)^{2}+(y + 4)^{2}=4\).

Step4: Identify center and radius

The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.
Here \(a = 3\), \(b=-4\), \(r = 2\).

Answer:

A. Center (3, -4); radius 2