Attachment 25055

Find the radius of the circle at the bottom of the image!!! i do need to get it done asap so if anyone could help please

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- Oct 5th 2012, 11:01 AMmidnightnurseSolve this please
Attachment 25055

Find the radius of the circle at the bottom of the image!!! i do need to get it done asap so if anyone could help please - Oct 5th 2012, 04:52 PMSorobanRe: Solve this please
Hello, midnightnurse!

We need more information.

Quote:

A full description of the figure is required.

Presently, I amthe following "facts" . . .*guessing*

We have a quadrant of a circle with radius 4 (the "4-circle").

There is a semicircle of radius 2 (the "2-circle").

The "4 cm" at the bottom isthe length of the chord.*not*

. . It is obviously a radius of the 4-circle.

We have a mystery circle tangent to the 2-circle.

It appears to extendthe 4-circle.*beyond*

With no further constraints, it could haveradius.*any*

- Oct 5th 2012, 06:10 PMjohnsomeoneRe: Solve this please
I assume the problem intends this:

P1) Rightmost semicircle is centered on that vertical line, has diameter 4 cm.

P2) Bottom shape is a semicircle whose center is on that horizontal line.

P3) The circles touch at a point (hence tangentially).

P4) The leftmost point of the bottom circle is 4 cm from the bottom-most point of the right semicircle.

In that case, you might set the problem up this way:

0) All distances in centimeters.

1) Bottom left point is the origin.

2) Bottom semicricle has radius r.

3) Right semicircle has radius 2 and center (4, 2).

Then by #2, bottom semicircle has center at (r, 0).

Here's the key observation: since they touch tangentially, the distance between two centers is the sum of their radii.

That becomes Distance( (r,0), (4,2) ) = 2 + r.

Thus:

$\displaystyle \sqrt{ (4-r)^2 + (2-0)^2} = 2+r$

$\displaystyle \sqrt{ (4-r)^2 + 4} = 2+r$

$\displaystyle (4-r)^2 + 4 = (2+r)^2$

$\displaystyle (16 - 2(4)(r) + r^2) + 4 = 4 + 2(2)(r) + r^2$

$\displaystyle r^2 - 8r + 20 = r^2 +4r + 4$

$\displaystyle - 8r + 20 = 4r + 4$

$\displaystyle - 8r -4r = 4 - 20$

$\displaystyle -12r = -16$

$\displaystyle r = \frac{-16}{-12}$

$\displaystyle r = \frac{4}{3}$.

The answer is that the bottom semicircle has a radius of $\displaystyle \frac{4}{3}$ cm. - Oct 5th 2012, 08:22 PMMaxJasperRe: Solve this please
http://mathhelpforum.com/attachment....1&d=1349497246

Radius of small tangent circle as a function of theta:

http://mathhelpforum.com/attachment....1&d=1349497258