Solving Common Chord of Two Circles

This article is inspired by one of the confusing problems in 2013 MMC for third year elimination round. The problem is all about the intersecting circles given the radii of two circles and the distance between their centers.

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,

intersecting circles

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

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.
Dan

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