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?

1 Answer
Jun 27, 2017

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]}