Graph the Circle
x^2+y^2+4x-2y-20=0
I get:
(x^2+4x)+(y^2-2y)=20
(x^2+4x+4)+(y^2-2y+1)=20+16+4
(x+2)^+(y-1)^2=(40/2)^2
But I don't think its right can you help me with what I did wrong? Thanks
(x^2+4x+4)+(y^2-2y+1)=20+16+4
Should be,
(x^2 +4x + 4)+(y^2 - 2y +1 )= 20 + 4 + 1
(x^2 +4x + 4)+(y^2 - 2y +1 )= 25
(x + 2)^2 + (y - 1)^2 = 25
Thanks, I see what I did, I squared when adding to the right side
To graph the equation of a circle, you need to rewrite it in the standard form, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Let's start with your equation: x^2 + y^2 + 4x - 2y - 20 = 0.
Step 1: Group the x-terms together and the y-terms together.
(x^2 + 4x) + (y^2 - 2y) - 20 = 0.
Step 2: Complete the square for the x-terms.
To complete the square for the x-terms, take half of the x-coefficient (which is 4 in this case), square it (16), and add it inside the parentheses. Don't forget to subtract it outside the parentheses to maintain the equation's balance.
(x^2 + 4x + 4) + (y^2 - 2y) - 20 - 4 = 0.
(x + 2)^2 + (y^2 - 2y) - 24 = 0.
Step 3: Complete the square for the y-terms.
To complete the square for the y-terms, take half of the y-coefficient (which is -2 in this case), square it (4), and add it inside the parentheses. Don't forget to subtract it outside the parentheses to maintain the equation's balance.
(x + 2)^2 + (y^2 - 2y + 1) - 24 - 1 = 0.
(x + 2)^2 + (y - 1)^2 - 25 = 0.
Now, the equation is in standard form: (x + 2)^2 + (y - 1)^2 = 25.
The center of the circle is at (-2, 1), and the radius is √25 = 5.
To graph the circle, plot the center (-2, 1) on the coordinate plane and draw a circle with a radius of 5 units around that point.