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

Group the \(x\) - terms and \(y\) - terms together: \((x^{2}+8x)+(y^{2}-6y)= - 21\)

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

For \(x^{2}+8x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = x\) and \(2b=8\) (so \(b = 4\)). Then \(x^{2}+8x=(x + 4)^{2}-16\)

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

For \(y^{2}-6y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\) and \(2b = 6\) (so \(b=3\)). Then \(y^{2}-6y=(y - 3)^{2}-9\)

Step4: Substitute back into the equation

Substitute into \((x^{2}+8x)+(y^{2}-6y)=-21\): \((x + 4)^{2}-16+(y - 3)^{2}-9=-21\)

Step5: Simplify the equation

Simplify to get the standard - form of the circle equation \((x - h)^{2}+(y - k)^{2}=r^{2}\). \((x + 4)^{2}+(y - 3)^{2}-25=-21\), then \((x + 4)^{2}+(y - 3)^{2}=4\)

Answer:

A. 2 units