# Two balls lie on the ground touching, if one ball has a radius 8 units and the POC is 10 units above the ground what is the radius of the ball.POC - Point of contact...

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I'm having trouble getting my figure to come out right, but I get a different answer. If you connect the centers of both circles with a straight line (which passes through the POC) and draw a horizontal line from the center of the smaller circle to the circle, another horizontal line from the POC to directly above the bottom of the larger circle, and also drop vertical lines from the POC and the center of the larger circle, you get two similar triangles.

The hypotenuse of the smaller triangle is 8, and the hypotenuse of the larger one is the unknown radius r. The vertical leg of the smaller one is 2, and the vertical leg of the larger one is r-10.

Then by similarity I get `8/2=r/(r-10),` so r=40/3.

If that's hard to follow or if I made a mistake, let me know.

Since both balls are lying on the ground we can imagine they are touching x-axis from above. And since we know radius of one ball we can write its equation. Equation of circle is:

`(x-a)^2+(y-b)^2=r^2`

where center of circle is in `(a,b)`

If we put `a=0` then `b=r=8`

So **equation of the first circle** (blue) is

`x^2+(y-8)^2=64`

**We can now calculate POC:**

Let's assume the second ball is right from the first. We know that `y` coordinate of POC is 10. By putting this in our equation we get

`x^2+(10-8)^2=64`

`x=2sqrt(15)`

`"POC"=(2sqrt(15),10)`

We can now calculate **tangent line at POC:**

Equation of tangent line at point `(x_0,y_0)` of circle `(x-a)^2+(y-b)^2=r^2`:

`(x_0-a)(x-a)+(y_0-b)(y-b)=r^2`

Thus we have

`sqrt(15)x+y=40`

This is also tangent of the second circle. Notice that x-axis is also tangent of second circle (because circle is on the ground and y-axis is our ground). So point `T` at which two tangents are intersected is equaly distant from POC and point `G` at which second circle is touching x-axis.

**Let's claculate** `T`:

Equation of x-axis is `y=0` thus

`sqrt(15)x+0=40`

`x=40/sqrt(15)=(8sqrt(15))/3`

`T=((8sqrt(15))/3,0)`

**Distance between** **POC** **and** `T`:

`d("POC",T)=sqrt(((8sqrt(15))/3-2sqrt(15))^2+10^2)=sqrt(110)`

**We can now claculate** `G:`

x coordinate of `G` is equal to sum of x coordinate of `T` and distance `d("POC",T)`. y coordinate of `G` is 0 because it is on x-axis (ground). Hence

`G=((8sqrt(15)+3sqrt(110))/3,0)`

**We can now calculate center os second circle:**

Center of second circle lies on intersection of line defined by center of first circle `C_1=(0,8)` and POC and line

`x=(8sqrt(15)+3sqrt(110))/3` **(1)**

Equation of line through 2 points is:

`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

Thus equation of line through `C_1` and POC is:

`y=sqrt(15)/15x+8` **(2)**

So intersection of (1) and (2) is:

`y=8+(8sqrt(15)+3sqrt(110))/(3sqrt(15))`

That is y coordinate of center os the second circle and that is also the radius of second circle because the circle is touching x-axis from above.` `

`r=8+(8sqrt(15)+3sqrt(110))/(3sqrt(15))`

No, you were absolutely right...