Question

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

Explanation:

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

$$(x^{2}+8x)+(y^{2}-6y)+21 = 0$$

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

For \(x^{2}+8x\), add \((\frac{8}{2})^{2}=16\).
$$(x^{2}+8x + 16)+(y^{2}-6y)+21=16$$

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

For \(y^{2}-6y\), add \((\frac{-6}{2})^{2}=9\).
$$(x^{2}+8x + 16)+(y^{2}-6y+9)+21=16 + 9$$

Step4: Write in standard circle form \((x - a)^{2}+(y - b)^{2}=r^{2}\)

$$(x + 4)^{2}+(y - 3)^{2}=4$$
Here \(r^{2}=4\), so \(r = 2\)

Answer:

2 units