Section 2.2

Definition 3 (Equation of Circle)

The equation of a circle with radius \(r\) and centered at \((0,0)\) is:

\[x^2+y^2=r^2\]

The equation of a circle with radius \(r\) and centered at \((h,k)\) is:

\[(x-h)^2+(y-k)^2=r^2\]

Completing the Square

Remember \((x+a)^2=x^2+2ax+a^2\) and \((x-a)^2=x^2-2ax+a^2\).

Example 18

Given the expression \(x^2+x+C\). Find the value \(C\) to complete the square.

Solution:

If \(C=-2\), then \(x^2+x-2=(x+2)(x-1)\) which does not complete the square. For this problem, we want to find the \(C\) values to make a perfect square polynomial.

If \(C=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) then

$$x^2+x+\frac{1}{4} = \left(x+\frac{1}{2}\right)^2

Remember when completing the square we want the leading coefficient to be one.

Example 19

Using completing the square, solve the following equation:

\[ 6x^2+13x+5=0 \]

Solution:

First, we will divide by \(6\) to both sides of the equation to get the leading coefficient to be one. We then get

\[x^2+\frac{13}{6} x +\frac{5}{6} = 0\]

Next, we will set up the equation for completing the square:

\[x^2 +\frac{13}{6} x + C = -\frac{5}{6} + C\]

Here, we want to find \(C\) that will complete the square.

\[C=\left(\frac{\frac{13}{6}}{2}\right)^2 = \left(\frac{13}{12}\right)^2=\frac{169}{144}\]

Substituting the \(C=\frac{169}{144}\) back into the equation and solving for \(x\) we have:

\[\begin{align*} x^2 +\frac{13}{6} x + C & = -\frac{5}{6} + C\\ x^2 +\frac{13}{6} x + \frac{169}{144} & = -\frac{5}{6} + \frac{169}{144}\\ \left(x+\frac{13}{12}\right)^2 & = \frac{49}{144} x+\frac{13}{12} & = \pm \sqrt{\frac{49}{144}}\\ x & = \frac{13}{12} \pm \frac{7}{12} \end{align*}\]

That is, the solution set is \(\{-\frac{1}{2},-\frac{5}{3}\}\).

General to Standard Form Circle Equation

Next, we will use the knowledge of completing the square to write \(x^2+y^2+Dx+Ey+F=0\) to \((x-h)^2+(y-k)^2=r^2\). This way we will be able to identify the circle’s center and radius.

Example 21

Given \(2x^2+2y^2+2x-6y=45\). Rewrite the equation in the form \((x-h)^2+(y-k)^2=r^2\) then find the center and radius of the resulting circle.

Solution:

First, we will divide both sides of the equation by \(2\).

\[ x^2 + y^2 + x - 3y = \frac{45}{2} \]

Next, we will set up the completing the square for the \(x\) expression and the \(y\) expression.

\[ x^2 + x + C_x + y^2 - 3y + C_y = \frac{45}{2} + C_x + C_y\]

Next, we will find \(C_x\) and \(C_y\) to complete the square.

\[ C_x = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

\[ C_y = \left(\frac{-3}{2}\right)^2 = \frac{9}{4} \]

Substituting \(C_x=\frac{1}{4}\) and \(C_y=\frac{9}{4}\) back into the equation we have

\[\begin{align*} x^2 + x + C_x + y^2 - 3y + C_y & = \frac{45}{2} + C_x + C_y\\ x^2 + x + \frac{1}{4} + y^2 - 3y + \frac{9}{4} & = \frac{45}{2} + \frac{1}{4} + \frac{9}{4}\\ \left(x+\frac{1}{2}\right)^2 + \left(y-\frac{3}{2}\right)^2 & = 25 \end{align*}\]

Therefore, the center of the graph is \((-\frac{1}{2},\frac{3}{2})\) and the radius is \(r=5\).

Remember the standard form is \((x-h)^2+(y-k)^2=r^2\). We can rewrite our result as follows to identify the center and radius clearly:

\[\begin{align*} \left(x+\frac{1}{2}\right)^2 + \left(y-\frac{3}{2}\right)^2 & = 25\\ \left(x-(-\frac{1}{2})\right)^2 + \left(y-\frac{3}{2}\right)^2 & = 5^2 \end{align*}\]

Example 21

Given \(x^2+y^2+6x+8y+9=0\). Rewrite the equation in the form \((x-h)^2+(y-k)^2=r^2\) then find the center and radius of the resulting circle.

Solution:

First, we will set up the completing the square:

\[x^2 + 6x + C_x + y^2 + 8y + C_y = -9 + C_x + C_y\]

Next, we will find \(C_x\) and \(C_y\).

\[C_x = \left(\frac{6}{2}\right)^2 = (3)^2 = 9\]

and

\[C_y=\left(\frac{8}{2}\right)^2 = (4)^2 = 16\]

Substituting those values we have:

\[\begin{align*} x^2 + 6x + 9 + y^2 + 8y + 16 & = -9 + 9 + 16\\ (x+3)^2 + (y+4)^2 & = 16 \end{align*}\]

Therefore, the center of the circle is \((-3,-4)\) and the radius is \(r=4\).