Precalculus - Circles

Brief Introduction to Conic Sections and Circles

• The figures are a group of curves called Conic Sections. They are formed from the intersection of a plane and right circular cones. Notice in the figures that they do not intersect the vertex of the cone. 
• Basically, there are four types of conic sections - Circle, Parabola, Ellipse, and Hyperbola.
• In general, conic sections are represented by second-degree equations. This is written as $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ where A and C are nonzero integers.
• Considering the general form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, if $B=0 and A=C$, then the graph of this equation represents a CIRCLE.
• A circle is a locus moving in a plane in such a way that its distance from a fixed point called the center remains constant. 
• A circle is a special kind of ellipse.
• A circle is formed when a cone is cut perpendicular to its axis.

The equation of the circle where the center is  $\left(0,0\right)$ is given by $x^2+y^2=r^2$ where $r=radius$. 

The equation of the circle whose center is at $C\left(h,k\right)$ is given by ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ where $r=radius$

The equation of the circle whose radius is equal to 1 and center at $\left(0,0\right)$ is given by $x^2+y^2=1$. This is called the Unit Circle.

Solved Examples

Example 1. Find the center and radius of the circle represented by $x^2-8x-6y+y^2-11=0$.

Solution:

Using completing the square, we can solve for the coordinates of the center $C\left(h,k\right)$ and the radius.

$x^2-8x+y^2-6y-11=0$

$x^2-8x+16+y^2-6y+9=11+16+9$

${\left(x-4\right)}^2+{\left(y-3\right)}^2=36$

Hence the center is at $C\left(4,3\right)$ with a radius of $r=\sqrt{10}$

 

Example 2. What is the equation of the circle whose center is at $C\left(5,7\right)$ and passes through $\left(2,8\right)$?

Solution:

 In this case, we need to solve for the radius using the distance formula:

$r^2={\left(5-2\right)}^2+{\left(7-8\right)}^2$

$r^2=9+1\to r=\sqrt{10}$

Thus, the equation of the circle is ${\left(x-5\right)}^2+{\left(y-7\right)}^2=10$.

 

Example 3. What is the equation of the circle that passes through the points $\left(-3,-9\right), \left(2,-4\right), and \left(6,-6\right)$?

Solution:

Use the general equation of the circle $A{x}^2+C{y}^2+Dx+Ey+F=0$ where A = C = 1.

This forms a linear system $\left\{\begin{array}{l}-3D-9E+F=-90\\ 2D-4E+F=-20\\ 6D-6E+F=-72\end{array}\right.$

Solving the system yields $D=-4, E=18, and F=60$

Substitute the obtained values to the general equation $x^2+y^2+Dx+Ey+F=0$.

Thus, the general equation of the circle is $x^2-4x+y^2+18y+60=0$.

The standard form equation is ${\left(x-2\right)}^2+{\left(y+9\right)}^2=25$.

More Problems involving Circles, Tangents, and Radical Axis

Example 1. What is the equation of the circle whose center is at $C\left(-3,-4\right)$ a tangent to the line $3x+4y=15$?

Solution:

We need to apply the formula of the distance of a point to a line:$d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}$. This is aimed to determine the distance between the line and the center.

$d=\frac{\left|3\left(-3\right)+4\left(-4\right)-15\right|}{\sqrt{25}}\to d=8$

Hence, the equation of the circle is ${\left(x+3\right)}^2+{\left(y+4\right)}^2=64$.

 

Example 2. Find the radical axis of the given circles $x^2+y^2=4 and {\left(x-2\right)}^2+{\left(y+9\right)}^2=25$.

Solution:

Subtract the two equations. The resulting linear equation is the radical axis.

$$\begin{gathered} x^2-4x+y^2+18y=-60 \\ -\left(x^2+y^2=4\right) \end{gathered}$$

The radical axis is $2x-9y=32$.

The graph of the two circles and their radical axis is shown below.

Cheat Sheet

• Be careful in dealing with the signs when the goal is to find the coordinates of the center when the equation of the circle is given.
• For instance, the circle ${\left(x+8\right)}^2+{\left(y-3\right)}^2=15$ has a center at $C\left(-8,3\right)$. 
• The circle ${\left(x-7\right)}^2+{\left(y-1\right)}^2=49$ has a center at $C\left(7,1\right)$.
• In finding the radius of the circle, always take the positive root. For instance, the circle ${\left(x+\frac{1}{2}\right)}^2+{\left(y-\frac{1}{4}\right)}^2=16$, then $r^2=16\to r=4$.
• The alternative formula to find the radius of the circle is $r^2=\frac{B^2+C^2-4AD}{4A^2}$.
• Use completing the square when transforming the general equation of a circle to its standard form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$. 

Blunder Areas

• In a circle, $A=C$ in the equation $A{x}^2+C{y}^2+Dx+Ey+F=0$.
• In an ellipse, $AC>0$ in the equation $A{x}^2+C{y}^2+Dx+Ey+F=0$. 
• When the equation of a conic section is given, always know the difference between a circle and an ellipse.
• The discriminant of a circle is given by $B^2-4AC<0$ provided that $A=C and B=0$. 
• Do not be confused with the discriminant of an ellipse since it is also $B^2-AC<0$. The condition in this case is $AC>0$.
• In solving for the equation of the circle when the three points are given, we use $A{x}^2+C{y}^2+Dx+Ey+F=0 where A=C=1$ to generate systems of linear equations with D, E, and F as variables.