Section 6-2: Equations
Definition of a Circle
is the set of all points in a plane equidistant from a fixed point called
the center point.
We can derive the equation directly from the distance formula. If
we place the center point on the origin point, the equation of a circle
with center point (0, 0) and radius r is:
x2 + y2
Look at the family of
A simple translation
of the circle equation becomes:
(x - h)2 +
(y - k)2 = r2
With center at (h, k)
and radius r.
Here are some examples:
1) Find the center, radius
and graph the equation: (x - 2)2
+ (y + 5)2 = 17
Solution: Center point : (2, -5),
The graph is below:
+ y2 - 8x + 4y - 8
= 0 Find center, radius and graph.
Solution: We need to put the equation
into the correct form. We will do this by completing the square!!
(x2 - 8x
) + (y2 + 4y
) = 8 Complete the square! See 1.6
(x2 - 8x +
16) + (y2
+ 4y + 4)
= 8 + 16 + 4
(x - 4)2
+ (y + 2)2 = 28
Now it's in the correct form!!
Center point (4, -2) with radius =
The graph is:
3) Find the intersection
of the line y = x - 1 and the circle x2
+ y2 = 25.
Solution: A line could intersect a circle
twice or once (if it is tangent) or not intersect at all. To find
the intersection use substitution. Replace y in the circle equation
with x - 1.
x2 + (x - 1)2
x2 + x2
- 2x + 1 = 25
2x2 - 2x - 24 = 0
x2 - x - 12 = 0
(x - 4)(x + 3) = 0
x = 4 or x = -3
Substituting back in to find y gives the following points:
(4, 3) and (-3, -4) If you check these points in both equations,
you will discover they make both true. The graph of the system is:
What would it mean if
the solutions were imaginary?
4) Sketch the graph
Solution: The above graph is part of
a circle. Why? Because it only includes the positive values
for y. Thus it is the top half of a circle with radius 4. It
is a semi-circle. By only studying this part of the circle, it makes
it a function. It now passes the vertical line test. Circles
are not functions! Here is the graph:
That's it for circles.
Let's head toward Ellipses!