# Circle: General Equation

In addition to the center-radius form, discussed under another topic and shown again here, the equation of a circle also has a general form.

GENERAL EQUATION:

Let C, D and E be constants.

${x}^{2}+{y}^{2}+Cx+Dy+E=0$

Let the center be (h, k) and the radius be r.

${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$

The center-radius form gives the center coordinates (h, k) and the radius r at a glance, whereas the general form does not provide easy access to this information.

Thus it is desirable to change the general form of the equation into the center-radius form to get this information. This is accomplished by completing the square.

 STEPS FOR CONVERTING THE GENERAL FORM INTO THE CENTER-RADIUS FORM: 1. Group the x- and y-terms on the left-hand side of the equation. 2. Move the constant term to the right-hand side. 3. Complete the square for the x- and y-groups. a. Divide the x-term coefficient by 2, square the result and add it to the x-group. b. Divide the y-term coefficient by 2, square the result and add the result to the y-group. 4. Add the same quantity to the right-hand side that was added to the left-hand side. 5. Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively.
Let's use the completing the square procedure in some examples:

Example 1: Find the center and radius of the circle by converting the general equation to the center-radius form by completing the square.

${x}^{2}+{y}^{2}+2x-6y-12=0$

 Step 1: Group the x- and y-terms on the left-hand side of the equation. ${x}^{2}+{y}^{2}+2x-6y-12=0$ $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)-12=0$ Step 2: Move the constant term to the right-hand side. $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)=12$ Step 3: Complete the square for the x- and y-groups. $\left({x}^{2}+2x\right)+\left({y}^{2}-6y\right)=12$ Complete the square for the x-group (x2 + 2x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{2}{2}=1;{1}^{2}=1$ Add the result to the x-group. (x2 + 2x + 1) Complete the square for the y-group (y2 - 6y) Take the coefficient of the y-term, divide by 2 and square the result. $\frac{-6}{2}=-3;{\left(-3\right)}^{2}=9$ Add the result to the y-group. (y2 - 6y + 9) Final result. $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=12+?$ Step 4: Add whatever was added to the left-hand side to the right-hand side. $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=12+1+9$ $\left({x}^{2}+2x+1\right)+\left({y}^{2}-6y+9\right)=22$ Step 5: Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively. (x + 1)2 + ${\left(y-3\right)}^{2}={\sqrt{22}}^{2}$ ${\left[x-\left(-1\right)\right]}^{2}+{\left(y-3\right)}^{2}={\sqrt{22}}^{2}$
Example 2: Find the center and radius of the circle by converting the general equation to the center-radius form by completing the square.

$3{x}^{2}+{y}^{2}-12x+8y-1=0$

 Step 1: Group the x- and y-terms on the left-hand side of the equation. In order to properly create squared terms, the coefficient of the x2-term must be one. As such we factor out a three from the x-group. $3{x}^{2}+{y}^{2}-12x+8y-1=0$ $\left(3{x}^{2}-12x\right)+\left({y}^{2}+8y\right)-1=0$ $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)-1=0$ Step 2: Move the constant term to the right-hand side. $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)=1$ Step 3: Complete the square for the x- and y-groups. $3\left({x}^{2}-4x\right)+\left({y}^{2}+8y\right)=1$ Complete the square for the x-group 3(x2 - 4x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{-4}{2}=-2;{\left(-2\right)}^{2}=4$ Add the result to the x-group. 3(x2 - 4x + 4) Complete the square for the y-group (y2 + 8y) Take the coefficient of the y-term, divide by 2 and square the result. $\frac{8}{2}=4;{4}^{2}=16$ Add the result to the y-group. (y2 + 8y + 16) Final result. $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=1+?$ Step 4: Add whatever was added to the left-hand side to the right-hand side. $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=1+3\left(4\right)+16$ $3\left({x}^{2}-4x+4\right)+\left({y}^{2}+8y+16\right)=29$ Step 5: Write the x-group, y-group and constant as perfect squares. Then ensure that the values of h and k are subtracted from x and y respectively. $3{\left(x-2\right)}^{2}+{\left(y+4\right)}^{2}={\sqrt{29}}^{2}$ $3{\left(x-2\right)}^{2}+{\left[y-\left(-4\right)\right]}^{2}={\sqrt{29}}^{2}$ Step 6: Identify the center and the radius. Center = (2, -4) Radius = $\sqrt{29}$

 Related Links: Math algebra Parabola: Standard Equation Ellipse: Standard Equation Pre Calculus

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