Proposed Problem

See complete Problem 277 at:

gogeometry.com/problem/p277_tangent_circle_common_external_tangent.htm

Level: High School, SAT Prep, College geometry

## Thursday, April 2, 2009

### Problem 277: Tangent Circles, Common External Tangent

Labels:
common tangent,
external,
geometric mean,
radius,
tangent

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Join BE and draw AF perpendicular to BE meeting the latter in F. Now DE=AF and in tr. ABF: AB^2=AF^2+BF^2 or BF^2= AF^2 = BF^2 or DE^2 = AF^2 - BF^2. But AF=a+b and BF=a-b. Hence, x^2 =DF^2=(a+b)^2-(a-b)^2 or x^2 =4ab or x = 2*(ab)^(1/2)

ReplyDeleteQED

Ajit: ajitathle@gmail.com

(a+b)^2-(b-a)^2=x^2

ReplyDeletea^2+2ab+b^2-(b^2-2ab+a^2)=x^2

4ab=x^2

√(4ab)=x

2√(ab)=x

Draw the tangent line common to the circles A and B at the point C. The tangent line meets DE at the point X .

ReplyDeleteWe have:

DX=CX=XE= y

And

x=2y

The points ACXD and BCXE are concylical. Therefore ∠CXD =∠CBE

If we draw the line AX it bisects the angle ∠CXD , and the line BX bisects ∠CBE

Therefore triangles ACX and XCB are similar and we get :

y/b=a/y

y^2=ab

(x/2)^2=ab

x^2=4ab

x=2(ab)^(1/2)