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.

$x^{2} + y^{2} + C x + D y + E = 0$

CENTER-RADIUS EQUATION:

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

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

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.

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

Step 1: Group the x- and y-terms on the left-hand side of the equation.	$x^{2} + y^{2} + 2 x - 6 y - 12 = 0$ $(x^{2} + 2 x) + (y^{2} - 6 y) - 12 = 0$
Step 2: Move the constant term to the right-hand side.	$(x^{2} + 2 x) + (y^{2} - 6 y) = 12$
Step 3: Complete the square for the x- and y-groups.	$(x^{2} + 2 x) + (y^{2} - 6 y) = 12$ 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. $\frac{- 6}{2} = - 3; {(- 3)}^{2} = 9$ Add the result to the y-group. (y² - 6y + 9) Final result. $(x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) = 12 + ?$
Step 4: Add whatever was added to the left-hand side to the right-hand side.	$(x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) = 12 + 1 + 9$ $(x^{2} + 2 x + 1) + (y^{2} - 6 y + 9) = 22$
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 + ${(y - 3)}^{2} = {\sqrt{22}}^{2}$ ${[x - (- 1)]}^{2} + {(y - 3)}^{2} = {\sqrt{22}}^{2}$

Example 2: Find the center and radius of the circle by converting the general equation to the center-radius form by completing the square.

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

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.	$3 x^{2} + y^{2} - 12 x + 8 y - 1 = 0$ $(3 x^{2} - 12 x) + (y^{2} + 8 y) - 1 = 0$ $3 (x^{2} - 4 x) + (y^{2} + 8 y) - 1 = 0$
Step 2: Move the constant term to the right-hand side.	$3 (x^{2} - 4 x) + (y^{2} + 8 y) = 1$
Step 3: Complete the square for the x- and y-groups.	$3 (x^{2} - 4 x) + (y^{2} + 8 y) = 1$ Complete the square for the x-group 3(x² - 4x) Take the coefficient of the x-term, divide by 2 and square the result. $\frac{- 4}{2} = - 2; {(- 2)}^{2} = 4$ 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. $3 (x^{2} - 4 x + 4) + (y^{2} + 8 y + 16) = 1 + ?$
Step 4: Add whatever was added to the left-hand side to the right-hand side.	$3 (x^{2} - 4 x + 4) + (y^{2} + 8 y + 16) = 1 + 3 (4) + 16$ $3 (x^{2} - 4 x + 4) + (y^{2} + 8 y + 16) = 29$
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.	$3 {(x - 2)}^{2} + {(y + 4)}^{2} = {\sqrt{29}}^{2}$ $3 {(x - 2)}^{2} + {[y - (- 4)]}^{2} = {\sqrt{29}}^{2}$
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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