Check?

Find the center and radius of the given circle

1. x^2+y^2+10y+21=0

= 5

Correct.

Radius is 5 at least. (1,5) is center

Howzat?

x^2 + (y+5)^2 = 4
center is at (0,5), radius is 2

To find the center and radius of a circle given its equation, you can follow these steps:

Step 1: Rewrite the equation in standard form. The standard form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle, and r is the radius.

Given equation: x^2 + y^2 + 10y + 21 = 0

To rewrite it in standard form, we need to complete the square for the y terms:

(x^2) + (y^2 + 10y) + 21 = 0

Step 2: Complete the square for the y terms. To do this, we need to take half of the coefficient of y, square it, and add it to both sides of the equation. In this case, half of 10 is 5, and (5)^2 is 25.

(x^2) + (y^2 + 10y + 25) + 21 - 25 = 0 + 25 - 25

(x^2) + (y^2 + 10y + 25) + 21 - 25 = 0

(x^2) + (y^2 + 10y + 25) - 4 = 0

Step 3: Rewrite the equation by factoring the square of the binomial.

(x^2) + (y + 5)^2 - 4 = 0

Now the equation is in standard form, (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius.

Comparing it to the standard form, we can deduce that the center of the circle is (-h, -k), and the radius is sqrt(r^2).

Therefore, the center of the given circle is (0, -5), and the radius is sqrt(4) = 2.