Circle: General Equation
GENERAL EQUATION:
Let C, D and E be constants.
${x}^{2}+{y}^{2}+Cx+Dy+E=0$
CENTER-RADIUS EQUATION:
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. |
${x}^{2}+{y}^{2}+2x-6y-12=0$
Step 1: Group the x- and y-terms on the left-hand side of the equation. |
${x}^{2}+{y}^{2}+2x-6y-12=0$ $({x}^{2}+2x)+({y}^{2}-6y)-12=0$ |
Step 2: Move the constant term to the right-hand side. |
$({x}^{2}+2x)+({y}^{2}-6y)=12$ |
Step 3: Complete the square for the x- and y-groups. |
$({x}^{2}+2x)+({y}^{2}-6y)=12$ Complete the square for the x-group (x^{2} + 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. (x^{2} + 2x + 1) Complete the square for the y-group (y^{2} - 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. (y^{2} - 6y + 9) Final result. $({x}^{2}+2x+1)+({y}^{2}-6y+9)=12+?$ |
Step 4: Add whatever was added to the left-hand side to the right-hand side. |
$({x}^{2}+2x+1)+({y}^{2}-6y+9)=12+1+9$ $({x}^{2}+2x+1)+({y}^{2}-6y+9)=22$ |
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} + ${\left(y-3\right)}^{2}={\sqrt{22}}^{2}$ ${\left[x-\left(-1\right)\right]}^{2}+{\left(y-3\right)}^{2}={\sqrt{22}}^{2}$ |
$3{x}^{2}+{y}^{2}-12x+8y-1=0$
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 x^{2}-term must be one. As such we factor out a three from the x-group. |
$3{x}^{2}+{y}^{2}-12x+8y-1=0$ $(3{x}^{2}-12x)+({y}^{2}+8y)-1=0$ $3({x}^{2}-4x)+({y}^{2}+8y)-1=0$ |
Step 2: Move the constant term to the right-hand side. |
$3({x}^{2}-4x)+({y}^{2}+8y)=1$ |
Step 3: Complete the square for the x- and y-groups. |
$3({x}^{2}-4x)+({y}^{2}+8y)=1$ Complete the square for the x-group 3(x^{2} - 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(x^{2} - 4x + 4) Complete the square for the y-group (y^{2} + 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. (y^{2} + 8y + 16) Final result. $3({x}^{2}-4x+4)+({y}^{2}+8y+16)=1+?$ |
Step 4: Add whatever was added to the left-hand side to the right-hand side. |
$3({x}^{2}-4x+4)+({y}^{2}+8y+16)=1+3\left(4\right)+16$ $3({x}^{2}-4x+4)+({y}^{2}+8y+16)=29$ |
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. |
$3{\left(x-2\right)}^{2}+{\left(y+4\right)}^{2}={\sqrt{29}}^{2}$ $3{\left(x-2\right)}^{2}+{\left[y-\left(-4\right)\right]}^{2}={\sqrt{29}}^{2}$ |
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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