Write the standard form of the equation of the circle x^2+y^2+4x-2y-20=0. Find the center and radius.
How do I do this? pleaaase help.
thank you so much! ;)
complete square:
(x+2)^2+(y-1)^2 = 20
Then minus 2^2+1^2 = 5, so add 5 to the 20
we get (x+2)^2+(y-1)^2 = 25
sqrt 25 = 5
so, 5 is the radius
the center is at (x+2)^2 = 0
(y-1)^2 = 0
so x = -2
y = 1
so the center is at (-2,1)
To find the standard form of the equation of the circle, you will need to complete the square for both the x and y terms.
First, let's rearrange the equation:
x^2 + 4x + y^2 - 2y = 20
Now, we will complete the square for the x terms:
x^2 + 4x + 4 + y^2 - 2y = 20 + 4
(x + 2)^2 + y^2 - 2y = 24
Next, we will complete the square for the y terms:
(x + 2)^2 + (y - 1)^2 - 1 = 24 + 1
(x + 2)^2 + (y - 1)^2 = 25
Now, we have the equation in the standard form for a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle and r represents the radius.
From our equation, we can determine that the center is (-2, 1) and the radius is √25, which simplifies to 5.
So, the standard form of the equation of the circle is (x - (-2))^2 + (y - 1)^2 = 5^2. The center of the circle is (-2, 1) and the radius is 5.
To write the standard form of the equation of a circle, x^2 + y^2 + 4x - 2y - 20 = 0, we need to complete the square for both the x and y terms.
1. Rearrange the equation by grouping the x and y terms together: (x^2 + 4x) + (y^2 - 2y) = 20.
2. For the x terms, take half the coefficient of x (which is 4), square it (which is 4^2 = 16), and add it to both sides of the equation: (x^2 + 4x + 16) + (y^2 - 2y) = 20 + 16.
3. Similarly, for the y terms, take half the coefficient of y (which is -2), square it (-2^2 = 4), and add it to both sides of the equation: (x^2 + 4x + 16) + (y^2 - 2y + 1) = 20 + 16 + 4.
4. Simplify the equation: (x + 2)^2 + (y - 1)^2 = 41.
Now we have the equation in standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Comparing it with our equation (x + 2)^2 + (y - 1)^2 = 41, we can see that the center of the circle is at (-2, 1), and the radius is sqrt(41).
So, the equation of the circle in standard form is (x + 2)^2 + (y - 1)^2 = 41, with center (-2, 1) and radius sqrt(41).