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 $x^{2} + y^{2} = 4$ $x^2 + y^2 =4$ and $x^{2} + y^{2} - 6 x + 2 = 0$ $x^2 +y^2 -6x +2 = 0$ ?

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Answer 1 · 2017-06-27T05:26:50

$2 \sqrt{3}$ $2sqrt3$ units

Here we have equations of two circles.

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

Circle [B]: $x^{2} + y^{2} - 6 x + 2 = 0$ $x^2+y^2-6x+2=0$

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 $x^{2} + y^{2} = 4$ $x^2+y^2 = 4$ from [A] directly into [B]:

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

$- 6 x = - 6 \to x = 1$ $-6x =-6 -> x=1$

From [A} $1 + y^{2} = 4$ $1+y^2 =4$

$y^{2} = 3 \to y = \pm \sqrt{3}$ $y^2 = 3 -> y=+-sqrt3$

Hence our points of intersection and hence the endpoint of the common chord are $(1, + \sqrt{3})$ $(1, +sqrt3)$ and $(1, - \sqrt{3})$ $(1, -sqrt3)$

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

$l = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$ $l = sqrt((x_1-x_2)^2 + (y_1-y_2)^2)$

Plugging in values of $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$

Chord length $= \sqrt{{(1 - 1)}^{2} + {(\sqrt{3} + \sqrt{3})}^{2}}$ $= sqrt((1-1)^2 + (sqrt3+sqrt3)^2)$

$= \sqrt{0 + {(2 \sqrt{3})}^{2}} = 2 \sqrt{3}$ $= sqrt(0+(2sqrt3)^2) = 2sqrt3$ 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]}

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 =4x2+y2=4x^2 + y^2 =4 and x^2 +y^2 -6x +2 = 0x2+y2−6x+2=0x^2 +y^2 -6x +2 = 0?