Equations of Circles — Challenge
A circle is every point a fixed distance from a fixed point. Write that with the distance formula and you have its equation.
What these problems cover
These questions use 6 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Radius — Centre to edge. Half the diameter.
- Diameter — All the way across through the centre. d = 2r.
- Distance formula — d = √(x2 - x1)2 + (y2 - y1)2.
- Midpoint — The point halfway between two others: average the x values and average the y values.
- Completing the square — Rewriting x2 + bx as a perfect square minus a constant.
- Simplest radical form — No perfect square factors left under the root.
Problem 1
What is the equation of the circle with centre (0,0) and radius 5?
- x2+y2=5
- x2+y2=25
- x2+y2=10
- (x-5)2+y2=0
- x+y=5
Worked answer
x2+y2=25
A circle centred at the origin has equation x2+y2 = r2.
With r=5 the right-hand side is 25, not 5.
Check with the point (5,0), which should be on the circle: 25+0 = 25.
Problem 2
What is the centre of the circle (x-3)2+(y+2)2=16?
- (3,-2)
- (-3,2)
- (3,2)
- (-3,-2)
- (2,3)
Worked answer
(3,-2)
The standard form is (x-h)2+(y-k)2 = r2, with centre (h,k).
Here x-3 gives h = 3, and y+2 is y-(-2), so k = -2.
The centre is (3,-2): each sign in the equation is the opposite of the coordinate.
Problem 3
What is the radius of the circle (x-1)2+(y-4)2=49?
- 49
- 14
- 7
- 24.5
- 4
Worked answer
7
The right-hand side of the standard form is r2, not r.
So r2 = 49 and r = √49 = 7.
Taking 49 itself is the commonest error here.
Problem 4
What is the equation of the circle with centre (2,-1) and radius 3?
- (x+2)2+(y-1)2=9
- (x-2)2+(y-1)2=9
- (x-2)2+(y+1)2=9
- (x-2)2+(y+1)2=3
- (x-2)2+(y+1)2=6
Worked answer
(x-2)2+(y+1)2=9
Substitute into (x-h)2+(y-k)2 = r2 with h=2, k=-1 and r=3.
y-(-1) becomes y+1, and r2 = 9.
So the equation is (x-2)2+(y+1)2 = 9.
Problem 5
What is the radius of the circle x2+y2=36?
- 36
- 6
- 18
- 12
- 9
Worked answer
6
The equation is x2+y2 = r2, so r2 = 36.
Taking the square root gives r = 6.
The diameter would be 12.
Problem 6
Does the point (3,4) lie on the circle x2+y2=25?
- Yes
- No
- Only if x=0
- It cannot be decided
- Only if y=0
Worked answer
Yes
Substitute the coordinates into the left-hand side: 32+42 = 9+16 = 25.
That equals the right-hand side, so the point satisfies the equation and lies on the circle.
This is the 3-4-5 right triangle: the point is exactly 5 units from the origin.
Problem 7
What is the centre of the circle x2+y2=100?
- (0,0)
- (10,10)
- (100,100)
- (10,0)
- (0,10)
Worked answer
(0,0)
There is nothing subtracted from x or y, so both h and k are zero.
The centre is therefore the origin, (0,0), and the radius is 10.
Problem 8
What is the diameter of the circle (x+5)2+(y-2)2=9?
- 3
- 6
- 9
- 18
- 4.5
Worked answer
6
From r2 = 9 the radius is 3.
The diameter is twice the radius: 2· 3 = 6.
Stopping at the radius, or at 9, are the two usual slips.