3. The Circle
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a. Circle Formulas
Center at the Origin
The circle with centre (0, 0) and radius r has the equation:
x2 + y2 = r2
This means any point (x, y) on the circle will give the radius squared when substituted into the circle equation.
Center not at the Origin
The circle with centre (h, k) and radius r has the equation:
(x − h)2 + (y − k)2 = r2
These formulas are a direct result of Pythagoras' Formula for the length of the hypotenuse of a right triangle.
Sketch the circle x2 + y2 = 4.
Find the center and radius first.
The equation is in the form x2 + y2 = r2, so we have a circle, center (0, 0) with radius `r=sqrt4=2`.
Sketch the circle (x − 2)2 + (y − 3)2 = 16
Find the center and radius first.
The equation is in the form (x − h)2 + (y − k)2 = r2, so we have a circle with centre at (2, 3) and the radius is `r=sqrt(16)=4`.
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Sketch the circle (x + 4)2 + (y − 5)2 = 36
The equation is in the form (x − h)2 + (y − k)2 = r2.
We have a circle with centre at (−4, 5) and the radius is `r=sqrt(36)=6`.
Be careful with positive and negative x- and y-values in this work!
b. The General Form of the Circle
An equation which can be written in the following form (with constants D, E, F) represents a circle:
x2 + y2 + Dx + Ey + F = 0
This is called the general form of the circle.
Find the centre and radius of the circle
x2 + y2 + 8x + 6y = 0
Sketch the circle.
Please revise Completing the Square first...
Our aim is to get the equation into the form: (x − h)2 + (y − k)2 = r2
We complete the square on the x-related portion and on the y-related portion, at the same time.
Group the x parts together and the y parts togther:
Complete the square on each of the x and y parts.
`(x^2+8x+16)+(y^2+6y+9)` `=16+9` `=25`
This is now in the format we require and we can determine the center and radius of the circle.
So the centre of the circle is (-4, -3) and the radius is 5 units.
Note that the circle passes through (0, 0). This is logical, since (−4)2 + (−3)2 = (5)2
1. Find the equation of the circle with centre `(3/2, -2)` and radius `5/2`.
Centre `(3/2, -2)` radius `5/2`.
General form of the equation of a circle:
The required equation for this case:
There is no need to expand this out, since this is the most useful form of the equation.
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2. Determine the centre and radius and then sketch the circle:
3x2 + 3y2 − 12x + 4 = 0
We complete the square as we did in an earlier example above.
First, we collect the x parts together and the y parts together, then divide throughout by 3.
Then we complete the square on the x part. We do not need to do so for the y part because there is no single y term (only a y2 term).
So the circle has centre `(2,0)` and has radius `sqrt(8/3)~~1.63`.
3. Find the points of intersection of the circle
x2 + y2 − x − 3y = 0
with the line
y = x − 1.
We solve the 2 equations simultaneously by substituting the expression
`y = x -1`
into the expression
[See some background to this at: Algebraic Solution of Systems of Equations.]
So we see that the solutions for x are `x = 1` or `x = 2`. This gives the corresponding y-values of `y = 0` and `y = 1`.
So the points of intersection are at: `(1, 0)` and `(2, 1)`.
We can see that our answer is correct in a sketch of the situation:
Exercise for You
Where did (0.5, 1.5) for the center of the circle come from?
Conic section: Circle
How can we obtain a circle from slicing a cone?
Each of the lines and curves in this chapter are conic sections, which means the curves are formed when we slice a cone at a certain angle.
If we slice a cone with a plane at right angles to the axis of the cone, the shape formed is a circle.