Proposed Problem

Click the figure below to see the complete problem 290.

See also:

Complete Problem 290

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Monday, May 18, 2009

### Problem 290: Internally Tangent circles, Radius, Perpendicular, Tangent

Labels:
circle,
perpendicular,
radius,
tangent

Subscribe to:
Post Comments (Atom)

Let A be(0,0) and B (r,h). The two circles are:

ReplyDeletex^2+y^2=a^2 and (x-r)^2+(y-h)^2=r^2. On solving the two equations simultaneously and making the discriminant zero to obtain the condition of the two circles touching each other, we get: -r^2a^4 +4r^4a^2+2(hra)^2-h^4r^2=o which, in turn, gives: 2ar= (a^2 - h^2). Now BE^2=(a+r)^2 + h^2 and EF^2 = BE^2 - r^2 =(a+r)^2+h^2 - r^2 = a^2+(a^2 -h^2) + h^2 = 2a^2. Or EF=V2*a or x = V2*a as depicted in the diagram.

Ajit: ajitathle@gmail.com

Let r the radius of the circle with center B.

ReplyDeleteBy cosine law EB^2=EA^2+AB^2-2EA.ABcos(EAB)=a^2+AB^2-2.a.ABcos(EAB), but since the circle with center A and radius a and the circle with radius r and center B y are tangents internally then AB=a-r , so EB^2=a^2+(a-r) ^2-2a(a-r)cos(EAB), now since EAB=90ยบ+DAB we got cos(EAB)=-sen(DAB), so EB^2=2a^2+r^2- 2ar+2a(a-r)sen(DAB), but sen(DAB)=r/(a-r), asi EB^2=2a^2+r^2-2ar+2ar=2a^2+r^2, Then by pitagoras theorem EF^2=EB^2-FB^2=2a^2+r^2-r^2=2a^2, which leads to EF^2=2a^2, QED.

AB = a-b so (a-b)^2 - b^2 + (a+b)^2 = x^2 + b^2 by applying Pythagoras to 3 right triangles BEF, BEG and ABG where G is the foot of the perpendicular to EA.

ReplyDeleteSimplifying x = a sqrt 2

Note that therefore EF = ED

Sumith Peiris

Moratuwa

Sri Lanka