#### Question

What is the radius of a circle whose equation is x^2+y^2+8x−6y+21=0?

Question

What is the radius of a circle whose equation is x^2+y^2+8x−6y+21=0?

To determine the radius of the circle given by the equation `x^2+y^2+8x−6y+21=0`, we need to rewrite the equation in standard form `(x-a)^2 + (y-b)^2 = r^2`, where `(a,b)` is the center of the circle and `r` is its radius.

To do this, we can complete the square for both `x` and `y` terms:

```

x^2 + 8x + y^2 - 6y + 21 = 0

(x^2 + 8x + 16) + (y^2 - 6y + 9) = -21 + 16 + 9 // adding and subtracting appropriate constants, completing the square

(x + 4)^2 + (y - 3)^2 = 2^2

```

So, the center of the circle is `(-4, 3)` and its radius is `4`.