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]}