what is the radius of a circle whose equation is $x^{2}+y^{2}-10x + 6y+18 = 0$?\n2 units\n4 units\n8…

Explanation:

Step1: Complete the square for (x) terms

For (x^{2}-10x), we have ((x - 5)^{2}=x^{2}-10x + 25).

Step2: Complete the square for (y) terms

For (y^{2}+6y), we have ((y + 3)^{2}=y^{2}+6y+9).

Step3: Rewrite the given equation

The original equation (x^{2}+y^{2}-10x + 6y+18 = 0) can be rewritten as ((x - 5)^{2}-25+(y + 3)^{2}-9+18=0). Simplify it: ((x - 5)^{2}+(y + 3)^{2}-16 = 0), then ((x - 5)^{2}+(y + 3)^{2}=16).

Step4: Find the radius

The standard form of a circle equation is ((x - a)^{2}+(y - b)^{2}=r^{2}), where (r) is the radius. Since (r^{2}=16), then (r = 4).

Answer:

4 units