Tangent Circles

Let the radius of smaller circle be $R$ image $900^2+(900-R)^2=(900+R)^2 \implies R=225$

Let larger radius be $R =900$ and smaller radius be $r$ .

Draw a line parallel to given line through center of smaller circle.

Draw a perpendicular on it from center of a larger circle.

We get a right angled triangle with sides, $R, R-r$ and $R+r$

By Pythagoras theorem, $R^2 = (R+r)^2 - (R-r)^2 \Rightarrow R = 4\cdot r \Rightarrow r = R / 4 = 900 / 4 = \boxed{225}$

Apply to Ford Circle's formula for 2 tangent circles and 1 circle between 2 circles same as fig. Ford Circle

$\frac{1}{\sqrt{R_{middle}}} = \frac{1}{\sqrt{R_{left}}} + \frac{1}{\sqrt{R_{right}}}$

$R_{left,right} = 900;$

$\frac{1}{\sqrt{R_{middle}}} = \frac{1}{\sqrt{900}} + \frac{1}{\sqrt{900}}$

$\frac{1}{\sqrt{R_{middle}}} = \frac{2}{30} = \frac{1}{15}$

Therefore: $R_{middle} = 15^{2} = \boxed{225}$

Draw a line from the center of the right big circle (radius = 900) to the center of the small circle, then draw a line perfectly horizontal until you are right under the center of the big circle, and then go back to the center of the big circle. This is a right triangle. Let's call the radius of small circle $r$ .

What you can see is that the horizontal side is, in fact, $900$ , the vertical side is $900 - r$ , and the hypotenuse is $900 + r$ . Now that we have that, all we must do is use Pythagorean theorem ( $a^2 + b^2 = c^2$ ) and solve for $r$ , the radius of the small circle.

$900^2 + (900 - r)^2 = (900 + r)^2$

$900^2 + (900^2 - 1800r + r^2) = 900^2 + 1800r + r^2$

$900^2 = 3600r$

$\frac{900^2}{3600} = r = \boxed{225}$

First we make some assumptions here

We can now draw out the trapezium as well as turning it into a triangle

Using Pythagoras Theorem

$(900+X)^{2}$ = $900^{2}$ + $(900-X)^{2}$

$X^{2}$ + 1800X + 810000 = 810000 + $X^{2}$ - 1800X + 810000

3600X = 810000

X = $\frac{810000}{3600}$ = 225

We see, from the Pythagorean Theorem, that

$(900)^2 + (900-r)^2 = (900+r)^2 \implies r = \boxed {225}.$

r = raio da circunferência menor --------> r = ?

R = raio da circunferência maior ---------> R = 900

• Resolvendo esta equação:

$(R+r)^{2}$ = $(R)^{2}$ + $(R-r)^{2}$ --------> $(900+r)^{2}$ = $(900)^{2}$ + $(900-r)^{2}$

r = $\boxed{225}$

Solution is belo w

x 2^2+(r 1-r 2)^2=(r 1+r_2)^2

for x_2, giving

x 2=2sqrt(r 1r_2).

x 3^2+(r 1-r 3)^2 = (r 1+r_3)^2

(x 3-x 2)^2+(r 2-r 3)^2 = (r 2+r 3)^2

for x 3 and r 3, giving

x 3 = (2r 1sqrt(r 2))/(sqrt(r 1)+sqrt(r_2))

r 3 = (r 1r 2)/((sqrt(r 1)+sqrt(r_2))^2).

The latter equation can be written in the form

1/(sqrt(r 3))=1/(sqrt(r 1))+1/(sqrt(r_2)).

It all depend upon the theorem of tangent to circle, two tangent meeting at a point have equal length. So we can solve by applying this principle, we will get r=900/4=225

In this specific formation : two touching congruent circles, with a smaller circle in between, that is tangent to both larger circles; the radius (r) of the smaller circle is always 1/4 of the radius (R) of the larger circles.

r = R/4 = 900/4 = 225

From seeing the figure. The radius should be 1/4 of the 1st circle.

diameter of small circle=900/2=450,
radii of small circle=450/2=225

Let $k_1$ be curvature of two big circles & $k_2$ be curvature of small circle. Here we can apply special case of descartes' theorem where one mutually tangent circle is a line & two circles are equal.

$k_2 = k_1+ k_1 + 2 \sqrt{k_1 k_1}$

$k_2 = \frac{1}{900} + \frac{1}{900} + 2 \sqrt{ \frac {1}{900} \cdot \frac{1}{900}}$

$k_2 = \frac{2}{900} + \frac{2}{900}$

$k_2 = \frac{1}{225}$

hence, the radius = $\frac{1}{curvature}$ = 225 .

If you join the middle of the circle to the middle of the smaller circle of radius r, and then draw a triangle with one side being 900 and other from center of smaller circle to the point where 2 bigger circles touch, you will get a right angled triangle with sides 900, 900 - r and hypotenuse of 900 + r. Applying pythagorean theorem, you get (900+r)^2 = (900-r)^2 + 900^2. Solving this for r you get r = 225

2*(diameter of small circle)=(radius of bigger circle)/2

use ford circle law 1/[sqrt{r}]radius of small circle=1/\sqrt{R}RADIUS OF LEFT CIRCLE+1/\sqrt{R'}RADIUS OF RIGHT CIRCLE

1/[sqrt{r}]=1/\sqrt{900}+1/\sqrt{900} 1/[sqrt{r}]=1/30+1/30

i think now you can do ur self.

Let the radius of the small circle = r ---> (900+r)^2 = 900^2 + (900-r)^2 solving ---> r = 225

in the following fig. name first and second circle wid centre as p and q.

centre of the smaller circle be O.

join O to q that would be r +R(radius of the smaller circle + radius of the bigger circle) likewise form a right angle triangle of sides, R+r,R, R-r inside the circle of centre q

so by Pythagoras theorem, r+R^{2} = R-r^{2} + R^{2} ......by this equation u will get r=225