G.GPE.A.1 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore
1
The Equation of a circle
The invention of Cartesian coordinates in the 17th century by René Descartes altered
mathematics greatly. It provided a systematic way to connect Euclidean geometry to algebra.
Using the Cartesian plane (also known as the coordinate plane) geometric shapes of all kinds
could be described as an equation – an algebraic equation involving the coordinates of the
points lying on the shape. For example, a circle with a radius 7 in a plane may be described as
the set of all points whose coordinates x and y satisfy the equation x2 + y2 = 49.
Cartesian coordinates are the foundation of analytic geometry and provide enlightening geometric
interpretations for many other branches of mathematics such as linear algebra, complex analysis, differential
geometry, multivariate calculus, group theory, and more.
A group of curves often analyzed using the Cartesian plane are the conic
sections. The conic sections refer to the cross sections of the cone or double
cone. The conic sections are the ellipse, circle, parabola, and hyperbola. A
circle cross section is formed when you cut a cone parallel to its circular base.
In this objective we will analyze circles and their algebraic equations.
The circle is defined as the locus (set) of points equidistant from a given point, called the center. When we
apply this definition to the coordinate plane we realize that the Pythagorean Theorem will help us greatly
because of its ability to calculate the distance of a segment. Actually the Pythagorean Theorem takes on a new
name when we move from the general plane to the coordinate plane, we call it the distance formula. Many
students use these two relationships regularly and never realize that they are actually the same relationship.
a 2  b2  c2
B (x2,y2)
 run    rise    distance 
2
2
2
 x2  x1    y2  y1    distance 
2
c
b
a
2
 x2  x1    y2  y1 
2
a 2  b2  c 2
2
2
(y2 - y1)
Rise
(x2 - x1)
A (x1,y1) Run
 distance
The most basic circle is the one centered at the origin (0,0). Using the
Pythagorean Theorem we can derive an equation that represents all the points
that are a distance of r, the radius, away from the origin.
a 2  b2  c2
The most basic circle equation is:
 run    rise   r
2
2
 x2  x1    y2  y1   r 2
2
2
 x  0   y  0  r 2
2
2
x2  y 2  r 2
2
x y  r
2
2
2
This is a circle centered at
the origin (0,0) with a radius of r.
P(x,y)
r
O(0,0)
G.GPE.A.1 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore
2
So x 2  y 2  9 is a circle centered at the origin with a radius of 3 cm and x 2  y 2  64 is a circle centered at
the origin with a radius of 8 cm. One of the errors that students often make is that they do not square root
the value when listing the radius value. Now not all circles are centered at the origin - wouldn’t that be easy.
Let us do this exercise again except that we will use a generic center, C (h,k).
Using the Pythagorean Theorem again we can derive an equation for all the
points that are a distance of r, the radius, away from the generic center, C(h,k).
a 2  b2  c2
 run    rise   r 2
2
2
 x2  x1    y2  y1   r 2
2
2
 x  h   y  k   r 2
2
2
P(x,y)
(y - k)
Thus the generic circle for all centers
and all radii is:
 x  h   y  k 
2
2
r
C(h,k)
(x - h)
2
This is a circle centered at C (h,k)
with a radius of r.
A few examples would be:
 x  2   y  5  100
2
 x  8  y 2  25
2
2
14   x  3   y  1
2
49  x 2   y  4 
2
2
Circle centered at (2,-5) with a radius of 10 cm.
Circle centered at (8,0) with a radius of 5 cm.
Circle centered at (-3,1) with a radius of 14 cm.
Circle centered at (0,4) with a radius of 7 cm.
1. Determine the equation of circle G, given the following information.
a) G(0,0) radius 4
b) G (3,7) radius 10
c) G(-3,5) diameter 6
____________________
____________________
____________________
d) G(-5,-1) radius 11
____________________
2. Determine the center and radius of the circle.
2
b) x2  ( y  8)2  4
a)  x  3  ( y  4)2 16
c)  x  2   ( y  1)2 100
d)  x  1  x 2 17
Center ( ____ , ____ )
Center ( ____ , ____ )
Center ( ____ , ____ )
Center ( ____ , ____ )
Radius ______
Radius ______
Radius ______
Radius ______
2
2
One of the errors students make is that in the equation they see  x  1 and they think the x value is -1 but it
2
is actually +1. Remember the general equation, r 2   x  h    y  k  , already has a negative value and so
2
2
when the h value is positive it would leave the equation in its negative form,  x  1 . Most students just
2
memorize that the value is the opposite of what they see in the equation.
r 2   x  h   y  k 
2
2
r 2   x  (1)    y  k 
2
r 2   x  1   y  k 
2
2
2
If the x value of the center was a negative, when substituted into the
equation it would transform the equation into a positive value. Students
don’t like this but when the value is positive, the equation shows a negative
form and when the value is negative, the equation shows a positive form.
Video: http://geometrycommoncore.screencasthost.com/gpea1ws1
2a. C (-3,4) r = 4
Answers:
1a.
b. C (0,-8) r = 2
b.
c. C (2,1) r = 10
c.
d. (1,0) r =
d.