Find the ratio of area of $K$ to that of $M$.

Question

Circle $K$ has diameter $AB$. Circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$. Circle $M$ is tangent to circle $K$, to circle $L$ and to $AB$. Find the ratio of area of $K$ to that of $M$.

What I Tried: Here is a picture :-

Mark all the points, do the labelling and draw perpendiculars and all, and in the end doing all these in Geogebra I found that $PN$ is almost equal to $\frac{1}{4}$th of $KB$'s length, where $KB$ is the radius of the large circle. That is possible only when $\Delta OQP \cong \Delta KQP$ , but I am really having a hard time showing this, and I got no idea how to proceed. I tried joining more lines and even doing Pythagorean Theorem but it had no use.

Can anyone help me? Thank You.

Source:- $1976$ AHSME Problem $24$ (Check here :- https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems)

Answer 1

Let $K(R)$, $L(R/2)$, $M(r)$.

$$PQ^2= OP^2 - OQ^2 = \color{blue}{PK^2} - QK^2$$

$$\Rightarrow (R/2+r)^2 - (R/2-r)^2 = \color{blue}{(R-r)^2} - r^2$$

$$\therefore \boxed{r=R/4}$$

Thinking method : Centers of two tangent circles are collinear with the tangency point. You did not use this condition for circles $K$ and $M$.

Answer 2

$$\begin{cases} x^2+y^2=(2-y)^2&\text{Pyth. th. in } \Delta KNP\\ x^2+(1-y)^2=(y+1)^2&\text{Pyth. th. in } \Delta POQ\\ \end{cases} $$ Solution is $x= \sqrt{2},y= \frac{1}{2}$

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