# Solving Common Chord of Two Circles

Problem:

Circles A and B with radii 5 and 7 respectively intersect at points M and N. If the distance between their centers is 6, find the measure of common chord MN?

Solution:

Using the Geogebra as drawing tool, below is the correct draft of the image,

MN is a chord of circle B. Since AB is a perpendicular bisector of MN, MO=ON.

Let x be the length of AO, then OB=6-x.

In $\triangle AOM$

$AM^2=AO^2+OM^2$

$5^2=x^2+OM^2$

$OM^2=25-x^2$

In $\triangle MOB$

$MB^2=OB^2+OM^2$

$7^2=(6-x)^2+OM^2$

$OM^2=49-(6-x)^2$

Equating  $OM^2$ in both triangles we have,

$OM^2=OM^2$

$25-x^2=49-(6-x)^2$

$25-x^2=49-(36-12x+x^2)$

$25-x^2=49-36+12x-x^2$

$12x=25+36-49$

$12x=12$

$x=1$

Solving for OM,

$OM^2=25-x^2$

$OM^2=25-1^2$

$OM^2=24$

$OM=2\sqrt{6}$

This means that  $MN=4\sqrt{6}$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.