1950's engineer book with "simple" geometry problem I can't solve

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mie

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See attached picture from a book. Unclear to me how to do this problem. Found in OLD science/engineering book from the 50's and was curious how to attack it (i apparently forgot geometry?

Here is my attempt to convert that picture, which as all the available information, into a word problem....in case you can't see the scan image from the book.

Circle1 with center at origin with radius= 1.5.
Circle2 sitting on X axis with diameter=D touching Circle1 at exactly one point.
Distance on X axis from Origin to the right-most-side of Circle2 (i.e. a perpendicular line to X axis that is tangent to Circle2) is 2.5
SO: Find "D"

I suspect it is trivial congruent triangles or something but it could require something more advanced. I don't see the answer.

Thanks for any help....
 

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Draw the following right triangle: The hypotenuse will go from the center of the left circle (or quarter circle if you insist) to the center of the right circle. This line will cross the point where the two circles intersect (convince yourself of that!!!) The 2nd leg of the triangle will go vertical from the center of the right circle down to the horizontal line in your diagram. Call this point C. Point C is also where the right circle intersect the horizontal line. The 3rd leg of the circle goes from point C to the center of the left circle.

Now the lengths of all three sides of this right triangle can be expressed in terms of D. Pythagoras' theorem should help from here.

Let us know how you make out. If you need further help please supply us with a labeled diagram including the right triangle so we can communicate better.
 
Denis said:
Carrying on from Jomo's:
A = center of larger circle
B = center of smaller circle
C = as per Jomo
Now you have right triangle ABC:
AB^2 = AC^2 + BC^2 ; ok?

R = radius larger circle = 3/2, r = radius smaller circle = ?
AB = R + r = 3/2 + r
AC = 5/2 - r
BC = r
So: (3/2 + r)^2 = (5/2 - r)^2 + r^2
This is a quadratic equation; solve for r: then D = 2r.

If you can't, then you need classroom help.
Click to expand...
NO No instead of r use D/2, then you can simply ignore the division by 2 in each term, not have to deal with fractions and in the end you will have D instead of R.
 
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Denis said:
Well, easier still:
u = 3/2, v = 5/2
(u + r)^2 = (v - r)^2 + r^2
Click to expand...
I still disagree. Once you solve you have to substitute back. Using D/2, when you solve for D you're done. I'm from good ole Brooklyn and never give in, especially when I'm correct.
 
Jomo said:
Draw the following right triangle: The hypotenuse will go from the center of the left circle (or quarter circle if you insist) to the center of the right circle. This line will cross the point where the two circles intersect (convince yourself of that!!!) The 2nd leg of the triangle will go vertical from the center of the right circle down to the horizontal line in your diagram. Call this point C. Point C is also where the right circle intersect the horizontal line. The 3rd leg of the circle goes from point C to the center of the left circle.

Now the lengths of all three sides of this right triangle can be expressed in terms of D. Pythagoras' theorem should help from here.

Let us know how you make out. If you need further help please supply us with a labeled diagram including the right triangle so we can communicate better.
Click to expand...
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Let the center of the big circle be O and center of the little circle be O'. let r = D/2

sin(O'OC) = (r)/(3/2+r)

Let the line OO' intersect the small circle at P (which is not on the big circle). I think you are implicitly assuming that a vertical line from P (down to x-axis) is tangent to the small circle and will touch the x-axis at a distance 2.5.

That assumption may not be true.
 
Subhotosh Khan said:
attachment.php

Let the center of the big circle be O and center of the little circle be O'. let r = D/2

sin(O'OC) = (r)/(3/2+r)

Let the line OO' intersect the small circle at P (which is not on the big circle).
Click to expand...
I am sorry but if two circles are tangent at one point, call this point P, then the line that joins the centers of the two circles will cross P.

Recall that if a line is tangent to a circle at point P, then the line from the center of the circle to P is perp. to the tangent line.
 
Jomo said:
I am sorry but if two circles are tangent at one point, call this point P, then the line that joins the centers of the two circles will cross P.

Recall that if a line is tangent to a circle at point P, then the line from the center of the circle to P is perp. to the tangent line.
Click to expand...

Going to the corner... misunderstood your "construction"

I thought you were extending the center-to-center line all the way to other side of the small circle.
 
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