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 Ax2+Bxy+Cy2+Dx+Ey+F=0 where A and C are nonzero integers.
  • Considering the general form Ax2+Bxy+Cy2+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  0,0 is given by x2+y2=r2 where r=radius

The equation of the circle whose center is at Ch,k is given by x-h2+y-k2=r2 where r=radius

The equation of the circle whose radius is equal to 1 and center at 0,0 is given by x2+y2=1. This is called the Unit Circle.

Solved Examples

Example 1. Find the center and radius of the circle represented by x2-8x-6y+y2-11=0.


Using completing the square, we can solve for the coordinates of the center Ch,k and the radius.




Hence the center is at C4,3 with a radius of r=10


Example 2. What is the equation of the circle whose center is at C5,7 and passes through 2,8?


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



Thus, the equation of the circle is x-52+y-72=10.


Example 3. What is the equation of the circle that passes through the points -3,-9, 2,-4, and 6,-6?


Use the general equation of the circle Ax2+Cy2+Dx+Ey+F=0 where A = C = 1.

This forms a linear system -3D-9E+F=-902D-4E+F=-206D-6E+F=-72

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

Substitute the obtained values to the general equation x2+y2+Dx+Ey+F=0.

Thus, the general equation of the circle is x2-4x+y2+18y+60=0.

The standard form equation is x-22+y+92=25.

More Problems involving Circles, Tangents, and Radical Axis

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


We need to apply the formula of the distance of a point to a line:d=Ax1+By1+CA2+B2. This is aimed to determine the distance between the line and the center.


Hence, the equation of the circle is x+32+y+42=64.


Example 2. Find the radical axis of the given circles x2+y2=4 and x-22+y+92=25.


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


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 x+82+y-32=15 has a center at C-8,3
  • The circle x-72+y-12=49 has a center at C7,1.
  • In finding the radius of the circle, always take the positive root. For instance, the circle x+122+y-142=16, then r2=16r=4.
  • The alternative formula to find the radius of the circle is r2=B2+C2-4AD4A2.
  • Use completing the square when transforming the general equation of a circle to its standard form x-h2+y-k2=r2

Blunder Areas

  • In a circle, A=C in the equation Ax2+Cy2+Dx+Ey+F=0.
  • In an ellipse, AC>0 in the equation Ax2+Cy2+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 B2-4AC<0 provided that A=C and B=0
  • Do not be confused with the discriminant of an ellipse since it is also B2-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 Ax2+Cy2+Dx+Ey+F=0 where A=C=1 to generate systems of linear equations with D, E, and F as variables.