what is the radius of a circle whose equation is $x^{2}+y^{2}+8x - 6y+21 = 0$?\n2 units\n3 units\n4 units\n5…

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for (x) and (y) terms. 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. For the (x) - terms: (x^{2}+8x=(x + 4)^{2}-16). For the (y) - terms: (y^{2}-6y=(y - 3)^{2}-9). The given equation (x^{2}+y^{2}+8x - 6y+21 = 0) can be rewritten as ((x + 4)^{2}-16+(y - 3)^{2}-9+21=0).

Step2: Simplify the equation

((x + 4)^{2}+(y - 3)^{2}-16-9 + 21=0), which simplifies to ((x + 4)^{2}+(y - 3)^{2}-4=0). Then ((x + 4)^{2}+(y - 3)^{2}=4).

Step3: Identify the radius

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

Answer:

2 units