Circles

Definition
The set of all points in a 2 dimensional surface all the same distance from one fixed point.

Calculate points

*Find the set of all points that are 2 units away from (0,0)

Length, radius or distance = Öx²+y²

Substitute: 2= Öx²+y²          4=x²+y²  
Equation of a circle centred (0,0)

*Write the equation of a circle centred in (0,0) and radius 5

r²=x²+y²

 
Substitute: 5²= x²+y²    25=x²+y²

Distance between any two points

*Find the distances between (-3,4) and (5,2)

If you imagine those two points, you will be able to imagine a right triangle.

This triangle measures 6, 8 and a length that we do not know yet.

Length=x²+y²

Substitute:  length=Ö6²+8²=Ö100=10

 Equation of a circle with centre other than (0,0)

*Write the equation of a circle centred in (2,3) and radius 5

length = ÖΔx²+Δy²

So,   r²= Δx²+Δy²

Substitute:  5²=Ö(x-2)²+(y-3)²         25= x-2)²+(y-3)²
 Points on, inside or outside the circle

*Given (x-2)² + (y+3)² = 25. Is the point (5,0) inside, outside or on the circle?

length = ÖΔx²+Δy²

Circle centre: (2,-3) and radius 5

Substitute: (5-2)² + (0+3)² = 18

18 < 25 So the point (5,0) is inside the circle.

-If the number were the same, it would be on the circle.

-If the number were larger, it would be outside the circle.
Points of intersecting

*Find the x and y points of intersection for this circle equation: (x-5)² + (y+4)² = 25

a)To find x point/s of intersection, let y=0, because this point will be in (x,0)

(x-5)² + (0+4)² = 25

(x-5)² = 9

Now there are two options (absolute value) because there are two points of intersection

X₁-5 = 3          x₁=8

X₂-5 = -3         x₂=2

X points of intersection are (2,0) and (8,0)

b) To find y point/s of intersection, let x=0

(0-5)² + (y+4)² = 25

(y+4)² = 0

Now there are not two options: 0 is no positive and no negative. There is only one point of intersection.

(y+4) = 0

y = -4

Y point of intersection is (0,-4)