James Gray

University of Edinburgh

Applicable Mathematics (Foundation) af0

5: Circles

5.1 Equation of a Circle

The equation of a circle centre `(0,0)` and radius `r` is `x^2 + y^2 = r^2`.

Reason:

We will show that the formula is true for a circle of radius 2.

By pythagoras, a point `P` lies on the circle precisely when

  • `x^2+y^2 = 2^2`.

That is when `x^2 + y^2=4`.

Equation of circle with centre `(a,b)` and radius `r` is `(x-a)^2 + (y-b)^2=r^2`.

Reason:

For the circle centre `(3,-1)` radius `5`, the point `P` lies on the circle if its distance from `(3,-1)` is `5`.  That is

  • `(x-3)^2+(y+1)^2=5^2`
  • so `(x-3)^2 + (y+1)^2 = 25`.

Note:

This is `(x-3)^2 + (y-(-1))^2 = 25` which is the boxed formula when `a=3` and `b=-1`.

Differet forms of circle equation

In Heinemann Ex 12F Q1(d) you found that the circle with centre `(-5,2)` and radius `sqrt(7)` has equation

Expand the brackets

What if we are given `x^2+10x + y^2-4y+22=0`?

To find the centre and radius we must complete the square

So the centre is `(-5,2)` and the radius is `sqrt(7)`.

5.2 Intersection of line and circle

3 cases

Example

Q: Find where the line `y=2x+8` meets the circle `x^2 + y^2 +4x +2y-20=0`.

A: Circle and line meet if coordinates satisfy both equations.  So substitute `y=2x+8` into the equation of a circle:

  • `x^2 + (2x+8)^2 +4x+2(2x+8)-20=0`

This can be rearranged and factorised to `5(x+2)(x+6)=0`.  (For the detail see Ex 10 on p216 of Heinemann.)  The solutions to this equation are `x=-2` and `x=-6`.

When `x=-2`, we have `y=2 times (-2) + 8 = 4`.

When `x=-6`, we have `y=2 times (-6) +8 = -4`.

So points of intersection are `(-2,4)` and `(-6,-4)`.

Example

Q: Show that the line `3x+y=-10` is tangent to the circle `x^2 + y^2 -8x +4y-20=0`.

A: Rearrange line equation to `y=-3x-10` and substitute into the circle equation.  This gives `10x^2+40x +40=0` (details of calculation given in Example 11 on p 217 of Heinemann).

The discriminant of `10x^2+40x+40=0` is `40^2-4 times 10 times 40 =0`.  So there is only one root of the quadratic.  Therefore there is only point of intersection and hence is tangent to the circle.

5.3 Tangents to a circle

Example

Q: Point `B(7,-3)` lies on circle `(x-2)^2 + (y+5)^2 = 29`.  Find the equation of tangent at `B`.

A: Centre of circle is `(2,-5)`.

  • `m_(CB) = (-3--5)/(7-2) = 2/5`

The tangent is perpendicular to `CB`.  Hence we have

  • `m = (-1)/(m_(CB))`

where `m` is the gradient of the tangent.  So `m = - 5/2`.

Since the point `(7,-3)` lies on the tangent we can find the equation of the tangent using the formula `y-b = m(x-a)`.  Therefore

  • `y-(-3) = - 5/2 (x-7)`
  • `y = -5/2 x + 35/2 -3`
  • `y= -5/2 x + 29/2.`