# Math Help - Solve this please

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

2. ## Re: Solve this please

Hello, midnightnurse!

Find the radius of the circle at the bottom of the image.

A full description of the figure is required.
Presently, I am guessing the following "facts" . . .

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 is not the length of the chord.
. . It is obviously a radius of the 4-circle.

We have a mystery circle tangent to the 2-circle.
It appears to extend beyond the 4-circle.

With no further constraints, it could have any radius.

3. ## Re: 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:

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

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

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

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

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

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

$- 8r -4r = 4 - 20$

$-12r = -16$

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

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

The answer is that the bottom semicircle has a radius of $\frac{4}{3}$ cm.

4. ## Re: Solve this please

Radius of small tangent circle as a function of theta: