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.