On a graph, x{\displaystyle x} tells you how many units from the center to go horizontally and y{\displaystyle y} tells you how many units from the center to go vertically to plot that point. Since h{\displaystyle h} is grouped with x{\displaystyle x} on the equation, it tells you how many units from the origin (that’s point 0,0{\displaystyle 0,0}) to go horizontally to plot the center of the circle. k{\displaystyle k} tells you how many vertical units from the origin to go to plot the center of the circle. When writing the equation of a circle, you always need to know the radius of the circle (r){\displaystyle (r)} and the coordinates of the center of the circle (h,k){\displaystyle (h,k)}.
For example, if you’re asked to find the center coordinates of a circle with the equation x2+(y−2)2=4{\displaystyle x^{2}+(y-2)^{2}=4}, you know the x{\displaystyle x} coordinate is 0{\displaystyle 0} because of the origin equation. The number grouped with y{\displaystyle y} is 2{\displaystyle 2}, so the center coordinates are (0,2){\displaystyle (0,2)}.
For example, if you’re given center coordinates of (4,6){\displaystyle (4,6)} and a radius of 3{\displaystyle 3}, your equation of a circle would be (x−4)2+(y−6)2=9{\displaystyle (x-4)^{2}+(y-6)^{2}=9}. Remember the formula is r2{\displaystyle r^{2}}, so you need to square 3{\displaystyle 3} for your final answer. If you’re given the diameter, remember to divide it in half because the form calls for the radius.
For example, if you have center coordinates (−4,−6){\displaystyle (-4,-6)} and radius 3{\displaystyle 3}, your equation of the circle would be (x+4)2+(y+6)2=9{\displaystyle (x+4)^{2}+(y+6)^{2}=9}. Likewise, if you were asked to find the center coordinates for the circle with the equation (x+1)2+(y+3)2=6{\displaystyle (x+1)^{2}+(y+3)^{2}=6}, you would know that the signs had changed, so the center coordinates must be negative: (−1,−3){\displaystyle (-1,-3)}.
For example, if you had 4x2+4y2−5x+8y−2=0{\displaystyle 4x^{2}+4y^{2}-5x+8y-2=0}, you would divide everything by 4{\displaystyle 4}, like so:(4x2)4+(4y2)4−5x4+8y4−24=04{\displaystyle {\frac {(4x^{2})}{4}}+{\frac {(4y^{2})}{4}}-{\frac {5x}{4}}+{\frac {8y}{4}}-{\frac {2}{4}}={\frac {0}{4}}}. You’d end up with x2+y2−54x+2y−12=0{\displaystyle x^{2}+y^{2}-{\frac {5}{4}}x+2y-{\frac {1}{2}}=0}. If the general form equation represents a circle, the two coefficients will always be the same number!
For example, x2+y2+14x−8y+56=0{\displaystyle x^{2}+y^{2}+14x-8y+56=0} would be x2+y2+14x−8y=−56{\displaystyle x^{2}+y^{2}+14x-8y=-56}.
For example, you would group x2+y2+14x−8y=−56{\displaystyle x^{2}+y^{2}+14x-8y=-56} as x2+14x+y2−8y=−56{\displaystyle x^{2}+14x+y^{2}-8y=-56}.
For example, if you have x2+14x{\displaystyle x^{2}+14x}, divide 14{\displaystyle 14} by 2{\displaystyle 2} to get 7{\displaystyle 7}. Then square 7{\displaystyle 7} to get 49.
For example, if you have y2−8y{\displaystyle y^{2}-8y}, your number would be 16: (−82)2=(−4)2=16{\displaystyle ({\frac {-8}{2}})^{2}=(-4)^{2}=16}.
For example, from x2+y2+14x−8y+56=0{\displaystyle x^{2}+y^{2}+14x-8y+56=0} you now have (x2+14x+49)+(y2−8y+16)=−56+49+16{\displaystyle (x^{2}+14x+49)+(y^{2}-8y+16)=-56+49+16}.
For example, for (x2+14x+49){\displaystyle (x^{2}+14x+49)}, you would get −14±142−4∗1∗492∗1=−7{\displaystyle {\frac {-14\pm {\sqrt {14^{2}-4149}}}{2*1}}=-7}. So your new parenthetical would be (x+7)2{\displaystyle (x+7)^{2}}. Do the same thing with (y2−8y+16){\displaystyle (y^{2}-8y+16)} to get (y−4)2{\displaystyle (y-4)^{2}}.
For example, if you’ve got (x+7)2+(y−4)2=−56+49+16{\displaystyle (x+7)^{2}+(y-4)^{2}=-56+49+16}, you’d add the three numbers on the right side to get (x+7)2+(y−4)2=9{\displaystyle (x+7)^{2}+(y-4)^{2}=9}. The number on the right side represents the radius squared. So if you’re asked for the radius, remember to take the square root of that number. In the example (x+7)2+(y−4)2=9{\displaystyle (x+7)^{2}+(y-4)^{2}=9}, it would be 3{\displaystyle 3}.
Hint: pay attention to the negative signs in front of the center coordinates.
Hint: look at the signs in the parentheses and compare them to the standard form for the equation.
Did you change the signs? Remember, two negatives make a positive, so the signs in the parentheses change to +{\displaystyle +}.
x2{\displaystyle x^{2}} and y2{\displaystyle y^{2}} already have a coefficient of 1{\displaystyle 1}, so you don’t have to do anything there. Group like terms: x2+4x+y2+8y−29=0{\displaystyle x^{2}+4x+y^{2}+8y-29=0} Move the constant to the right side of the equation: x2+4x+y2+8y=29{\displaystyle x^{2}+4x+y^{2}+8y=29} Get the number you need to complete the square for the x{\displaystyle x} terms: x=(42)2=(2)2=4{\displaystyle x=({\frac {4}{2}})^{2}=(2)^{2}=4} Get the number you need to complete the square for the y{\displaystyle y} terms: y=(82)2=(4)2=16{\displaystyle y=({\frac {8}{2}})^{2}=(4)^{2}=16} Plug your numbers in to both sides of the equation: (x2+4x+4)+(y2+8y+16)=29+4+16{\displaystyle (x^{2}+4x+4)+(y^{2}+8y+16)=29+4+16} Do the quadratic equation for the x{\displaystyle x} terms: −4±42−4∗1∗44∗1=−2{\displaystyle {\frac {-4\pm {\sqrt {4^{2}-414}}}{41}}=-2} Do the quadratic equation for the y{\displaystyle y} terms: −8±82−4∗1∗168∗1=−4{\displaystyle {\frac {-8\pm {\sqrt {8^{2}-4116}}}{81}}=-4} Plug your parenthetical expressions into the standard form: (x+2)2+(y+4)2=29+4+16{\displaystyle (x+2)^{2}+(y+4)^{2}=29+4+16} Add the numbers on the right side of the equation: (x+2)2+(y+4)2=49{\displaystyle (x+2)^{2}+(y+4)^{2}=49} Take the square root of 49{\displaystyle 49} to find the radius.
Why? Because you’re going to add 16{\displaystyle 16} to −20{\displaystyle -20}. Since 16{\displaystyle 16} is smaller than 20{\displaystyle 20}, you know the result is going to be negative. But a radius can’t be negative—so that’s it. You’ve solved the problem. Great work!
