find the center and radii of the circle 3x^2-3y^2+4x-5y+2=0?
In the first place, x^2-y^2+... is not a circle. But, given that you know it's an hyperbola, just complete the squares to find its standard form.
3x^2-3y^2+4x-5y+2=0
3x^2+4x - 3y^2-5y = -2
x^2 + 4/3 x - y^2 - 5/3 y = -2/3
(x^2 + 4/3 x + (2/3)^2) - (y^2 + 5/3 y + (5/6)^2) = -2/3 + (2/3)^2 - (5/6)^2
(x + 2/3)^2 - (y + 5/6)^2 = -11/12
(y + 5/6)^2 / (11/12) - (x + 2/3)^2 / (11/12) = 1
To find the center and radii of the circle, we need to rewrite the given equation in standard form, which is (x-h)^2 + (y-k)^2 = r^2.
The equation of the circle is:
3x^2 - 3y^2 + 4x - 5y + 2 = 0
First, let's group the x-terms and the y-terms:
(3x^2 + 4x) - (3y^2 + 5y) + 2 = 0
Next, complete the square for both the x-terms and the y-terms:
For the x-terms:
Take half of the coefficient of x, which is 4, and square it. (4/2)^2 = 2^2 = 4
Add and subtract 4 inside the parentheses:
(3x^2 + 4x + 4 - 4) - (3y^2 + 5y) + 2 = 0
Rearrange the terms:
(3x^2 + 4x + 4) - (3y^2 + 5y) + 2 - 4= 0
Simplify:
[(3x^2 + 4x + 4) - (3y^2 + 5y)] - 2 = 0
[(3x^2 + 4x + 4) - 3(y^2 + (5/3)y)] - 2 = 0
For the y-terms:
Take half of the coefficient of y, which is 5/3, and square it. (5/3 * 1/2)^2 = (5/6)^2 = 25/36
Add and subtract 25/36 inside the parentheses:
[(3x^2 + 4x + 4) - 3(y^2 + (5/3)y + 25/36)] - 2 + (3 * 25/36) = 0
Simplify:
[(3x^2 + 4x + 4) - 3(y^2 + (5/3)y + 25/36)] - 2 + 25/12 = 0
[(3x^2 + 4x + 4) - 3(y^2 + (5/3)y + 25/36)] + 13/12 = 0
Now, we can write the equation in standard form:
[(3x^2 + 4x + 4) - 3(y^2 + (5/3)y + 25/36)] + 13/12 = 0
Divide the entire equation by the coefficient of x^2 and y^2, which is 3:
[(x^2 + (4/3)x + 4/3) - (y^2 + (5/3)y + 25/36/3)] + 13/12/3 = 0
[(x^2 + (4/3)x + 4/3) - (y^2 + (5/3)y + 25/108)] + 13/36 = 0
Now, we can rewrite the equation in standard form:
(x + 2/3)^2 - (y + 5/6)^2 + 13/36 = 0
From the standard form, we can identify the center and radii of the circle:
Center: The center of the circle is (-h, -k), which is (-2/3, -5/6).
Radii: The radii of the circle can be found by taking the square root of the number on the right side of the equation, which is √(13/36).
Therefore, the center of the circle is (-2/3, -5/6), and the radii of the circle are √(13/36), which can be simplified to √13/6.
To find the center and radii of a circle given its equation, we need to rewrite the equation in a specific form—known as the standard form of a circle equation—which 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.
Let's begin by rearranging the given equation:
3x^2 - 3y^2 + 4x - 5y + 2 = 0
To complete the square for the x terms, we factor out a common factor from the x^2 and x terms:
3(x^2 + (4/3)x) - 3y^2 - 5y + 2 = 0
Next, we need to figure out what value to add to complete the square for the x terms. To find this value, we take half the coefficient of x (in this case, 4/3) and square it:
(4/3)/2 = 2/3
(2/3)^2 = 4/9
Now, we add this value to both sides of the equation:
3(x^2 + (4/3)x + 4/9) - 3y^2 - 5y + 2 + 4/3 = 4/9
Next, we complete the square for the y terms by applying the same process:
3(x^2 + (4/3)x + 4/9) - 3(y^2 + (5/3)y) + 2 + 4/3 = 4/9
To complete the square for the y terms, we take half the coefficient of y (in this case, 5/3) and square it:
(5/3)/2 = 5/6
(5/6)^2 = 25/36
We then add this value to both sides of the equation:
3(x^2 + (4/3)x + 4/9) - 3(y^2 + (5/3)y + 25/36) + 2 + 4/3 + 25/36 = 4/9 + 25/36
Simplifying:
3(x^2 + (4/3)x + 4/9) - 3(y^2 + (5/3)y + 25/36) + 83/36 = 49/36
Now, we can rewrite the equation as a standard form of a circle equation:
3(x + 2/3)^2 - 3(y + 5/6)^2 + 83/36 = 49/36
Comparing this to the standard form, we have:
(x + 2/3)^2 / (49/108) - (y + 5/6)^2 / (49/108) = 1
From this equation, we can identify the center of the circle as (-2/3, -5/6) and the radii are the square roots of (49/108) in both directions. So, the radii of this circle are sqrt(49/108) and -sqrt(49/108).
Therefore, the center of the circle is (-2/3, -5/6), and the radii are sqrt(49/108) and -sqrt(49/108).