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Problem 289: Tangent circles, Radius, Perpendicular, Tangent

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Saturday, May 9, 2009

### Problem 289: Tangent circles, Radius, Perpendicular, Tangent

Labels:
circle,
perpendicular,
radius,
tangent

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Let A be (0,0) and AD the x-axis, the radius of circle B = r. Hence B:(((a^2+2ar)^(1/2),r) and E:(0,a). Let F be (x1,y1). Thus we have

ReplyDeletex1-(a^2+2ra)^(1/2))^2+(y1-r)^2=r^2 since F lies on circle B. Considering slopes of EF & BF we get:-x1/(y1-a)=(y1-r)/(x1-(a^2+2ra)^(1/2)) and

(EF)^2 = (y1-a)^2 + x1^2. On solving the equations obtained we get: EF=V2*a or x=V2*a

Ajit: ajitathle@gmail.com

From triangle BFE we can say that: x^2=BE^2-r^2 if r is the radius of the circle B. But B is ((a^2+2ar)^(1/2),r)as may easily ascertained and A is (0,a). Hence we've: x^2 = [(a^2+2ar)^(1/2)-0]^2+(r-a)^2 - r^2 using the distance formula. This easily gives x^2 = 2a^2 w/o having to worry about the co-ords. of F.

ReplyDeleteAjit

Let radius of circle B be b.

ReplyDeleteAB=a+b

BD=b

AD^2 = (a+b)^2 - b^2 = a*a + 2ab

Now draw a line parallel to AD from B to AE which touches AE at say M. Now ME = a-b

EB^2 = (a-b)^2 + a^2 +2ab = 2*a*a + b*b

Now EF^2 + b^2 = EB^2

EF^2 = 2a*a + b*b -b*b = 2a*a

EF = a*sqrt(2).