what is the radius of a circle whose equation is x² + y²+8x - 6y + 21 = 0?\n2 units\n3 units\n4 units\n5 units

Explanation:

Step1: Rewrite the equation in standard form

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}+8x - 6y+21 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}+8x=(x + 4)^{2}-16$. For the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$. So the equation becomes $(x + 4)^{2}-16+(y - 3)^{2}-9+21 = 0$.

Step2: Simplify the equation

$(x + 4)^{2}+(y - 3)^{2}-16-9 + 21=0$. $(x + 4)^{2}+(y - 3)^{2}-4 = 0$. $(x + 4)^{2}+(y - 3)^{2}=4$.

Step3: Identify the radius

Since the standard - form is $(x + 4)^{2}+(y - 3)^{2}=2^{2}$, the radius $r = 2$.

Answer:

2 units