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: Complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}+8x)+(y^{2}-6y)+21 = 0$.
For the x - terms $x^{2}+8x$, add and subtract $(\frac{8}{2})^{2}=16$ inside the parentheses: $(x^{2}+8x + 16-16)+(y^{2}-6y)+21 = 0$, which can be rewritten as $(x + 4)^{2}-16+(y^{2}-6y)+21 = 0$.

Step2: Complete the square for y - terms

For the y - terms $y^{2}-6y$, add and subtract $(\frac{-6}{2})^{2}=9$ inside the parentheses: $(x + 4)^{2}-16+(y^{2}-6y + 9-9)+21 = 0$.
This becomes $(x + 4)^{2}-16+(y - 3)^{2}-9+21 = 0$.

Step3: Rewrite the equation in standard form

Simplify the left - hand side of the equation: $(x + 4)^{2}+(y - 3)^{2}-16-9 + 21=0$.
$(x + 4)^{2}+(y - 3)^{2}-4 = 0$.
Then $(x + 4)^{2}+(y - 3)^{2}=4$.
The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and r is the radius.

Step4: Determine the radius

Comparing $(x + 4)^{2}+(y - 3)^{2}=4$ with $(x - a)^{2}+(y - b)^{2}=r^{2}$, we have $r^{2}=4$, so $r = 2$.

Answer:

2 units