Section 2.2 - Circles

Definition 5

A circle is the set of all points in a plane that lie a given distance from a given point. The given distance is called the circle’s radius, denoted \(r\), and the given point is called the center, denoted \((h,k)\). The equation that satisfies this definition is

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

The equation \(x^{2}+y^{2}=1\) defines the unit circle (since the radius of the circle is \(1\) and is centered at \((0,0)\)).

Consider:

\[\begin{split}\begin{aligned} (x-h)^{2}+(y-k)^{2} & =c\iff\\ (x-h)^{2}+(y-k)^{2}-c & =0\iff\\ y^{2}-2\,k\,y+x^{2}-2\,h\,x+k^{2}+h^{2}-c & =0\iff\\ x^{2}+y^{2}+(-2h)x+(-2k)y+\left(k^{2}+h^{2}-c\right) & =0\iff\\ x^{2}+y^{2}+Dx+Ey+F & =0\end{aligned}\end{split}\]

Definition 6 (General Form of the Equation of a Circle)

From some real number \(D,\)\(E\), and \(F\), the equation

\[x^{2}+y^{2}+Dx+Ey+F=0\]

can have a graph that is a circle or a point or is nonexistent.

Corollary 1

Given

\[(x-h)^{2}+(y-k)^{2}=c\]

for any number \(c\).

1. If \(c>0\), then \(r^{2}=c\), and the graph of the equation is a circle with radius \(\sqrt{c}\).

2. If \(c=0\), then the graph of the equation is the single point \((h,k)\).

3. If \(c<0\), then no points satisfy the equation, and the graph is nonexistent.

Example 26

Solve by completing the square

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

Solution:

First, divide by \(6\) to get the leading coefficient equal to one.

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

Second, subtract \(\frac{5}{6}\) from both sides and leave space to “complete the square”.

\[x^{2}+\frac{13}{6}x+\underline{\,\,\,\,}=-\frac{5}{6}+\underline{\,\,\,\,}.\]

Third, complete the square by adding \(\left(\dfrac{\frac{13}{6}}{2}\right)^{2}=\left(\dfrac{13}{12}\right)^{2}=\dfrac{169}{144}\) to both sides

\[\begin{split}\begin{aligned} x^{2}+\frac{13}{6}x+\frac{169}{144} & =-\frac{5}{6}+\frac{169}{144}\\ \left(x+\frac{13}{12}\right)^{2} & =\frac{49}{144}\\ x+\frac{13}{12} & =\pm\sqrt{\frac{49}{144}}\\ x & =-\frac{13}{12}\pm\frac{7}{12}=\begin{cases} -\frac{13}{12}+\frac{7}{12} & =-\frac{6}{12}=-\frac{1}{2}\\ -\frac{13}{12}-\frac{7}{12} & =-\frac{20}{12}=-\frac{5}{3} \end{cases}\end{aligned}\end{split}\]

The solution \(x=\{-\frac{1}{2},-\frac{5}{3}\}\).

Example 27

Give the center center and radius of the circle with equation

\[2x^{2}+2y^{2}+2x-6y=45\]

Solution:

First, we will divide both sides by \(2\) to get \(x^{2}\) and \(y^{2}\) to have coefficients of one (in preparation of completing the square).

\[\begin{split}\begin{aligned} 2x^{2}+2y^{2}+2x-6y & =45\\ x^{2}+y^{2}+x-3y & =\frac{45}{2}\end{aligned}\end{split}\]

Second, we group up the \(x\) and \(y\) variable terms. Third, find the completing the square term by computing \(\left(\frac{b}{2}\right)^{2}\). For \(x\), \(b=1\). For \(y\), \(b=-3\).

\[\begin{split}\begin{aligned} \left(x^{2}+x+\underline{\,\,\,}_{1}\right)+\left(y^{2}-3x+\underline{\,\,\,}_{2}\right) & =\frac{45}{2}+\underline{\,\,\,}_{1}+\underline{\,\,\,}_{2}\\ \underline{\,\,\,}_{1} & =\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\\ \underline{\,\,\,}_{2} & =\left(\frac{-3}{2}\right)^{2}=\frac{9}{4}\\ \left(x^{2}+x+\frac{1}{4}\right)+\left(y^{2}-3x+\frac{9}{4}\right) & =\frac{45}{2}+\frac{1}{4}+\frac{9}{4}\\ \left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2} & =25\\ \left(x-\left(-\frac{1}{2}\right)\right)^{2}+\left(y-\frac{3}{2}\right)^{2} & =5^{2}\end{aligned}\end{split}\]

This means, the implicit equation \(2x^{2}+2y^{2}+2x-6y=45\) can be rewritten as \(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=25\). From the second equation we notice

\[\begin{split}\begin{aligned} (h,k) & =(-\frac{1}{2},\frac{3}{2})\\ r^{2} & =25\implies r=\sqrt{25}=5.\end{aligned}\end{split}\]

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

Example 28

Give the center and radius of the circle represented by the equation

\[x^{2}+y^{2}+6x+8y+9=0.\]

\[\begin{split}\begin{aligned} x^{2}+y^{2}+6x+8y & =-9\\ \left(x^{2}+6x+c_{1}\right)+\left(y^{2}+8y+c_{2}\right) & =-9+c_{1}+c_{2}\\ c_{1} & =\left(\frac{6}{2}\right)^{2}=\left(3\right)^{2}=9\\ c_{2} & =\left(\frac{8}{2}\right)^{2}=\left(4\right)^{2}=16\\ \left(x^{2}+6x+9\right)+\left(y^{2}+8y+16\right) & =-9+9+16\\ (x+3)^{2}+(y+4)^{2} & =16\\ (x-(-3))^{2}+(y-(-4))^{2} & =4^{2}\\ (h,k) & =(-3,-4)\\ r & =4\end{aligned}\end{split}\]