Isaiah's Math Lessons  Circles in Coordinate Geometry

Circle Definition:The locus of a point that moves such that its distance from a fixed point called the center is constant. The constant distance is called the radius, r of the circle.
General Equation:
Ax^{2} + Ay^{2} + Dx + Ey + F = 0
Or
x^{2} + y^{2} + Dx + Ey + F = 0Standard Equation:
Circle with center at (h, k) : (x−h)^{2} + (y−k)^{2} = r^{2}
Circle with center at (0, 0) : x^{2} + y^{2} = r^{2}NOTE: The r variable here stands for the radius of the circle.
Examples and Detailed Solutions:
1. What is the general equation of a circle with center at (1, 2) and radius equal to 3?
Based from the Standard equation of the Circle, we will have the below equation:
(x−h)^{2} + (y−k)^{2} = r^{2}
But (h, k) is (1,2) and r = 3
(x−1)^{2} + (y−(−2))^{2} = 3^{2}
(x−1)^{2} + (y+2)^{2} = 3^{2}
We know that (a+b)^{2} = a^{2} + 2ab + b^{2}
x^{2} − 2x + 1 + y^{2 }+ 4y + 4 = 9
Rearranging the Equation, we will have x^{2} + y^{2}  2x + 4y – 4 = 02. Find the equation in standard from, center and radius of a circle with equation 4x^{2}+ 4y^{2}−4x+ 12y−15=0
The trick here is to first make the coefficient of x^{2} and y^{2} be equal to 1.
Therefore, we divide the whole equation by 4. We divide it by 4 since that is the coefficient of both x^{2} and y^{2}.
(4x^{2}+ 4y^{2}−4x+ 12y−15) / 4 = 0 = > x^{2}+ y^{2}− x+3y− 15/4 = 0After making the coefficient of x2 and y2 equal to 1, we now group similar terms then completing the square
( x2 − x + _ ) + ( y2+ 3y+ _ ) = 15/4 + _ + _
How do we complete a square? First is we divide the coefficient of the middle term by 2 then square it. For the left side of the equation, the coefficient of the middle term is 1 which is the coefficient of x. We then divide 1 by 2 and then square it which results to 1/4. Same thing goes for the y terms. We divided 3 by 2 then square it which will result to 9/4.( x^{2}− x + 1/4) + ( y^{2} + 3y + 9/4) = 15/4 + 1/4 + 9/4
We then factor it to a binomial raised to the 2nd power. We copy the middle term of the expressions above.
(x−1/2)^{2} + (y+3/2)^{2} = ( 5/2)^{2}Center (1/2, 3/2) and r = 5/2
Additional Shortcut: How to find the Center of a Circle in General form having the equation
x^{2} + y^{2} + Dx + Ey + F = 0
Shortcut Formula (Without Completing the Square)
Center (−D/2 , E/2)
Radius = 1/2√(D^{2}+ E^{2} − 4F)Practice Problems:
1. What is the center of the circle with equation x^{2}+ y^{2}+ 4x−6y+9=0?
2. For what value of K will make x^{2}+ y^{2}+ 4x−6y+K=0 a point circle? (Clue : a point circle is a circle with radius = 0 )
3. Find the equation of the circle tangent to the xaxis with center at (1, 2)?
4. What is the value of the eccentricity of a circle? What does eccentricity means?Answers:
1. C (2,3)
2. K = 13
3. (x−1)^{2} + (y−2)^{2} = 2^{2}
4. e=0, it is defined as the parameter which is associated with every conic section. It can be thought as a measure of how much the conic deviates to become a circular figure. Other values of eccentricity are Parabola (e=1), Ellipse (0 < e < 1) and Hyperbola (e > 1).