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.
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.
2 answers
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,...)
1 answer