ahmed hisham project

1. Equation of Circle
Additional Mathematics

2. Distance between Two Points
1 2 3 4 5
-1
-2
-3
-4
-5
x
1
2
3
4
5
-1
-2
-3
-4
-5
y
By Phytagoras Theorem:

3. Equation of Circle with Centre (0, 0)
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y
Distance
What can you conclude from this pattern?

4. Equation of Circle with Centre (0, 0)
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y By applying Pythagoras theorem,
Distance
As the centre of the circle (0, 0)
and the radius OD, then the
equation of the circle:

5. Find an equation of the circle with center at (0, 0) that
passes through the point (-1, -4).
2
2
2
r
y
x
2
2
2
4
1 r
The point (-1, -4) is on the circle then
we may substitute the coordinate on
the equation to find the radius of the
circle
17
16
1
17
2
2
y
x
Since the center is at (0, 0) we'll have
Subbing this in for r2 we have:
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y

6. Equation of Circle with Centre (a, b)
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y
Distance

7. Equation of Circle with Centre (a, b)
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y
By applying distance formula,
Distance
As the centre of the circle (a, b)
and the radius OD, then the
equation of the circle:

8. Find an equation of the circle with center at (2, -1) and
radius 4.
Substitute the center and the radius into
the general equation
Expand and simply the equation to
give alternative form: 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
x
1
2
3
4
5
6
-1
-2
-3
-4
-5
-6
y
2 -1 4