Question

what is the radius of a circle whose equation is ( x^{2}+y^{2}+8x - 6y + 21 = 0 )?
3 units
2 units
4 units
5 units

Explanation:

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

Group \(x\) terms: \(x^{2}+8x=(x + 4)^{2}-16\)

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

Group \(y\) terms: \(y^{2}-6y=(y - 3)^{2}-9\)

Step3: Rewrite the circle equation

Substitute into original equation: \((x + 4)^{2}-16+(y - 3)^{2}-9+21 = 0\)
Simplify: \((x + 4)^{2}+(y - 3)^{2}-4 = 0\)
Then \((x + 4)^{2}+(y - 3)^{2}=4\)

Step4: Identify the radius

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

Answer:

2 units