Hello math bugs(π) and hivers(π)

I hope you had a good weekend.

I have come up with another cute one. The problem today is when a circle is inscribed a circle, you have to find out its radius when a non-parallel side is given. You may think, " No! there no side given, actually two portions of two sides given only!!"Actually if you check properly you will find the given side is hidden a bit.

Now if you find the side, give it a try first.Find the radius(r).

Okay, before heading towards solution, I just want to remind you some concepts/postulates/axioms first. Here we go:

β (1) The distance between two parallel tangents of a circle always passed through the centre.Thus the distance becomes its diameter.

β (2) The radius of a circle is always perpendicular to its tangent. Hence, in the problem, the diameter is perpendicular to both of the tangents. Check the following figure:

β
(3) Two tangents of a circle from a particular point outside cover equal distance to the point of contact.Check the following figure:

The same way the following figure is made:

β (4) Conditions of similarity of right angle triangles.And then need to find out thr relation between sides using it.If we combine the previous two figure we can find a right angle triangle.we just need to connect CO and DO. Check it below:

If you haven't figure out the answer yet, try it now ;you have all the concepts you need to solve it. And if you still don't get it, no worry ,let me now do it step by step.

You may be aware about similarity very well and if not check my previous detailed post on similarity.

Let's consider the right angle βCOD only. Using similarity we can write OPΒ² = CP ΓDP. How!! Check the following figure:

So, the required answer is "r" = square root of (4β2 Γ2β2) cm.That is "r" = 4 cm π. Check the figure also:

All the figures made by me using Android apps. There may be some mistake while typing or making figures, please excuse me for that.

Thank you so much friends for stoping by. I hope you enjoyed it. Have a good day.

All is well

Regards: @meta007