Question

A chord of a circle is a line segment whose endpoints are on the circle. Find the length of the common chord of the two circles whose equations are `x^2 + y^2 =4` and `x^2 +y^2 -6x +2 = 0`?

Answer 1

`2sqrt3` units

Explanation:

Here we have equations of two circles.

Circle [A}: `x^2+y^2 = 4`

Circle [B]: `x^2+y^2-6x+2=0`

To find the coordinates of the common chord we need to find the points of intersection of the [A] and [B]. i.e where equation [A] equals equation [B]

NB: In this special case we can replace `x^2+y^2 = 4` from [A] directly into [B]:

Thus: `4 -6x +2 =0`

`-6x =-6 -> x=1`

From [A} `1+y^2 =4`

`y^2 = 3 -> y=+-sqrt3`

Hence our points of intersection and hence the endpoint of the common chord are `(1, +sqrt3)` and `(1, -sqrt3)`

To find the length between these two points we use the formula:

`l = sqrt((x_1-x_2)^2 + (y_1-y_2)^2)`

Plugging in values of `(x_1,y_1)` and `(x_2,y_2)`

Chord length `= sqrt((1-1)^2 + (sqrt3+sqrt3)^2)`

`= sqrt(0+(2sqrt3)^2) = 2sqrt3` units

We can see the points of intersection and deduce the length of the common chord graphically below:

graph{(x^2+y^2-4)(x^2+y^2-6x+2)=0 [-6.243, 6.243, -3.12, 3.123]}