General Form for the Equation of a Circle

We now know the center-radius form for the equation of a circle centered at (h,k) and with a radius r is LaTeX: \left(x-h\right)^2+\left(y-k\right)^2=r^2.(x−h)2+(y−k)2=r2. If we square the binomials on the left side of this equation, we will get a polynomial which will have LaTeX: x^2,\:y^2,\:x,\:y\:x2,y2,x,yand number terms. The LaTeX: x^2\:x2and LaTeX: y^2\:y2terms will have a coefficient of 1. Thus, if we subtract LaTeX: r^2r2 from both sides of the equation and insert D and E in to represent the coefficients of the LaTeX: x\:xand LaTeX: y\:yterms and F to represent the number term, we get:

General Form of the Equation of a Circle

LaTeX: x^2+y^2+Dx+Ey+F=0x2+y2+Dx+Ey+F=0

It turns out this equation can have a graph that is a circle, or a point, or is nonexistant. To determine which of these graphs a particular equation in this form has, we must complete the square to obtain an equation similar to the center-radius form with LaTeX: \left(x-h\right)^2+\left(y-k\right)^2=c.(x−h)2+(y−k)2=c.

There are three possibilities for the graph of this equation based on the value of LaTeX: c.\:c.

  1.  If LaTeX: c>0,\:c>0,then LaTeX: r^2=c,\:r2=c,and the graph of the equation is a circle with radius LaTeX: \sqrt[]{c}.√c.
  2. If LaTeX: c=0,\:c=0,then the graph is a single point LaTeX: \left(h,k\right).(h,k).(One can think of this as a circle with radius 0)
  3. If LaTeX: c<0,\:c<0,then no points satisfy the equation, and the graph is nonexistant.

Example

Give the center and radius of the circle represented by the equation LaTeX: x^2+y^2+8x-6y+16=0x2+y2+8x−6y+16=0.

Solution

To tell what the graph would look like, we first have to complete the square. Below in red are the steps for completing the square, and in black are the steps for this particular example:

  • Add or subtract the same amount from both sides of the equation in order to get the number term to only be on the right, and all terms with either variable in them to the left of the equal sign.

We first need to subtract 16 from both sides of the equation, which gives us:LaTeX: x^2+y^2+8x-6y=-16x2+y2+8x−6y=−16

  • Now write the x terms in descending order next to each other with parentheses around them and the y terms in descending order next to each other with parentheses around them.

LaTeX: \left(x^2+8x\right)+\left(y^2-6y\right)=-16(x2+8x)+(y2−6y)=−16

  • To complete the square for each variable, take half of the coefficient of the first degree term, square this amount, and add the result to both sides of the equation.

For the x terms, the first degree term is LaTeX: 8x\:8xwhich has a coefficient of 8.  Half of 8 is 4, and 4 squared is 16, so to complete the square for the x's, we will add 16 to both sides of the equation:

LaTeX: \left(x^2+8x+16\right)+\left(y^2-6y\right)=-16+16(x2+8x+16)+(y2−6y)=−16+16

Similarly for the y's, we take half of -6, which is -3, square this, and we get 9. So, we need to add 9 to both sides of the equation to complete the square for the y's.

LaTeX: \left(x^2+8x+16\right)+\left(y^2-6y+9\right)=-16+16+9(x2+8x+16)+(y2−6y+9)=−16+16+9

  • Now rewrite the perfect square trinomials on the left as binomials squared, and combine the number terms on the right.

LaTeX: \left(x+4\right)^2+\left(y-3\right)^2=9(x+4)2+(y−3)2=9 or LaTeX: \left(x-\left(-4\right)\right)^2+\left(y-3\right)^2=3^2(x−(−4))2+(y−3)2=32

This is now in the form LaTeX: \left(x-h\right)^2+\left(y-k\right)^2=c\:(x−h)2+(y−k)2=cwith LaTeX: c=9\:c=9so LaTeX: c>0.c>0.

In fact, this equation is a circle in center-radius form with center LaTeX: \left(-4,3\right)(−4,3) and radius LaTeX: r=\sqrt[]{9}=3.r=√9=3.

When one is finished completing the square,

  • If the number term on the right is negative, then the graph is nonexistant.
  • If the number term on the right is zero, then the graph is the point (h,k), because the radius is 0.

Next complete Homework 2.2 on the next page.