How do I find the center and radius of this equation? x^2+y^2+6y+2=0
I don't know how to complete the square for x^2.
The standard form of the circle is
(x-h)²+(y-k)²=r*sup2;
where (h,k) is the centre of the circle
and r is the radius.
For a given equation of the circle
x² + y² + 2hx + 2ky + C = 0
1. examine the coefficient of the terms containing x² and y². They should be equal. If they are not equal in value and in sign, or if one of the terms is missing, the equation does not represent a circle.
2. If the coefficients of the terms x² and y² are not equal to +1, divide all the terms of the equation by the coeffieient to reduce them to 1.
3. Examine the coefficient of the x term. If it is zero, the centre of the circle lies along the y-axis. If not, complete the square by combining the x² and x terms into one single term by replacing x+2hx by (x+h)²-h².
4. Repeat 3 for the y-terms.
5. Transpose C and h² nad k² to the right hand side of the equal sign to get h²+k²-C.
To illustrate how to transform a general equation to the standard form, I will take a more general case (#89)
x²+y²-4x+10y+13=0
x²-4x + y²+10y +13 = 0
(x-2)²-2² + (y+5)²-5² + 13 = 0
(x-2)² + (y+5)² = 2²+5²-13
(x-2)² + (y+5)² = 4²
Therefore, by comparison with the standard form, we conclude that the centre is at (2,-5) and the radius is 4.
Complete the posted problem and post the answer for verification if necessary.
To complete the square of the term x² where the x-term is missing, it is equivalent to the term (x-0)², and consequently the centre is at (0,...)
To find the center and radius of the equation x^2 + y^2 + 6y + 2 = 0, we need to rewrite the equation in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center and r represents the radius.
To proceed, we can first rearrange the equation by moving the constants to the right side: x^2 + y^2 + 6y = -2.
Now, let's complete the square for the y terms. Since the coefficient of y^2 is 1, we can divide the coefficient of y by 2, square it, and add it inside the parentheses. In this case, the coefficient of y is 6, so we have y^2 + 6y. Dividing 6 by 2 gives us 3, and squaring it gives us 9. To keep the equation balanced, we need to add 9 inside the parentheses and subtract 9 on the right side:
x^2 + y^2 + 6y + 9 = -2 + 9.
Now, group the terms involving y:
x^2 + (y^2 + 6y + 9) = 7.
Next, rewrite the group involving y as a perfect square:
x^2 + (y + 3)^2 = 7.
Now, we can compare this expression with the standard form of a circle, (x - h)^2 + (y - k)^2 = r^2. We see that the center of the circle is (h, k) = (0, -3), and the radius squared, r^2, is equal to 7. Thus, the center of the circle is (0, -3), and the radius is √7.