find the radius and co-ordinates of the center of the circle: 2x^2+2y^2+x+y=o
2x2 + x + 2y2 + y = 0
2(x2 + x/2) + 2(y2 + y/2) = 0
2(x2 + x/2 + 1/4) + 2(y2 + y/2 + 1/4) - 2(1/4) - 2(1/4) = 0
2(x - 1/2)2 + 2(y - 1/2)2 = 1
(x - 1/2)2 + (y - 1/2)2 = 1/2
Center = (1/2,1/2) radius = 1/√2
Rats. Botched it.
2(x2 + x/2 + 1/16) + 2(y2 + y/2 + y/16) - 4(1/16) = 0
2(x + 1/4)2 + 2(y + 1/4)2 = 1/4
(x + 1/4)2 + (y + 1/4)2 = 1/8
center = (-1/4, -1/4)
radius = 1/√8
To find the radius and coordinates of the center of the circle given the equation 2x^2 + 2y^2 + x + y = 0, we first need to rewrite the equation in a specific form called the standard form equation for a circle.
The standard form equation for a circle is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.
Let's go step by step to rewrite the given equation in the standard form:
2x^2 + 2y^2 + x + y = 0
First, let's group the terms with x and y:
(2x^2 + x) + (2y^2 + y) = 0
Next, we can complete the square for both x and y individually.
For the x terms:
We want to add and subtract (1/2)^2 = 1/4 to complete the square. Adding 1/4 inside the parentheses will help us complete the square for the x terms.
(2x^2 + x + 1/4) - 1/4 + (2y^2 + y) = 0
We'll do the same for the y terms:
(2x^2 + x + 1/4) + (2y^2 + y + 1/4) - 1/4 - 1/4 = 0
Now, let's factor the x^2 + x + 1/4 and y^2 + y + 1/4 terms:
((sqrt(2)x + 1/2)^2) + ((sqrt(2)y + 1/2)^2) - 1/2 = 0
Finally, rearranging the equation, we have:
((sqrt(2)x + 1/2)^2) + ((sqrt(2)y + 1/2)^2) = 1/2
Comparing this equation with the standard form equation for a circle, we can see that the center coordinates are:
(h, k) = (-1/2, -1/2)
And the radius of the circle is given by the equation:
r^2 = 1/2
r = sqrt(1/2)
Therefore, the radius of the circle is (1/2)sqrt(2), and the center coordinates are (-1/2, -1/2).