Ascending Crescent

Let $R$ be the radius of the larger circle and $r$ be the radius of the smaller one. If $O'$ is the center of the smaller circle, then

1. $2r+4cm=2R$ #look at line $AB$
2. $(R-3cm)^{2}+(r-(R-4cm))^{2}=r^{2}$ #look at triangle $O'QO$

From this 2 equations, we get

$(R-3)^{2}+(r-(R-4))^{2}=r^{2}$ , $R=r+2$

$(r-1)^{2}+(2)^{2}=r^{2}$

$r^{2}-2r+1+4=r^{2}$

$2r=5$

Finally, we get that $\boxed{r=2.5cm}$ .

Let x be the radius of larger circle , then From triangle OBQ line QB = x2 + (x-3)2

and from triangle OAQ Line QA = (x + x - 4)2 - PQ2

as QB = QA , so x2 + (x-3)2 = (x + x - 4)2 - PQ2

x2 + x2 + 9 - 6x = (2x - 4)2 -PQ2

2x2 + 9 - 6x = 4x2 + 16 - 16x - PQ2

PQ2 = 4x2 - 2x2 + 16 - 9 - 16x + 6x

PQ2 = 2x2 + 7 -10x ------------------ (1)

Now from traingle PQB

PQ2 = (x-4)2 + (x-3)2

PQ2 = 2x2 - 14x + 25 ----------------------(2)

From Equation (1) and (2)

7 - 10x = 25 - 14x

4x = 18

x = 18 / 4

If r is the radius of small circle , then 2r = 2x - 4

2r = 2*18/4 - 4

2r = 9 - 4

r = 2.5

Let $R$ be the radius of the larger circle. Let $r$ be the radius of the smaller circle and $E$ it's center. let $\theta$ be the angle $\angle OEQ$

find a formula for the length line $AB$ in terms of $R$ and $r$ :

$2R=4+2r$

similarly for $OD$ :

$R=3+r\sin(\theta)$

and $OB$ :

$R=r+r\cos(\theta$

divide equation one by 2, cancel R from equations leaving two equations in terms of $\theta$ and $r$ :

$2+r=3+r\sin(\theta)$

$2+r=r+r\cos(\theta)$

rearrange for $\sin$ and $\cos$ :

$\sin(\theta)=\frac{r-1}{r}$

$\cos(\theta)=\frac{2}{r}$

now recall $\sin^2(\theta)+cos^2(\theta)=1$ so:

$(\frac{r-1}{r})^2+(\frac{2}{r})^2=1$

after algebra gives:

$2r=5$

so $r=2.5cm$

Let OP= x , OQ= y and O ' is the center of the smaller circle with a radius of r . now , 3+y=4+x or, y =x+1 then , 4+x =r -x +r or, r=x+2 again , at right angled triangle O ' PQ , using Pythagoras theorem , (r -x)^2 +y^2=r^2 or , 2rx=x^2+y^2 or, 2(x+2)x =x^ 2+ (x+1)^2 or, 2x^2+4x=x^2 +x^2+2x+1 or, 2x =1 or, x=0.5 so , r =x+2 = 0.5 + 2 = 2.5 cm :-)

PB=5cm then 5/2=2.5cm

By observation I can closely guess that the radius of the larger circle is about 4.5 cm. The smaller circle's diameter is larger than the larger circle's radius by 0.5 cm. Half of 5.0 is 2.5.

QJ is a chord of the small circle, where O is the medium point. By power of point, we get (QO)×(OJ) = (PO)×(OB), which is the same of (QO)^2 = (PO)×(OB). If we call the radius of the big circle of x, the expression will be (x-3)^2 = (x-4)x. Evaluating, the radius of the big one will be 4,5. That means the diameter of the small circle is x(radius of the big circle) + 0.5 = 5. Then, the radius of the small circle will be 2,5.