Prove that a line BP will pass through the centre O of the circle inscribed in a quadrant ABC Consider the figure as shown below:

Here, $ABC $ is a quadrant, a circle with centre $ O $ is inscribed in the quadrant such that it intersects the quadrant at $M,N $ and $P$;
Now,how can we prove that the straight line $BP$ will pass through the centre $O$ ?   
My Idea:
If we can prove that $\angle POB =180°$ ,then we have proved that $BP $ will pass through the centre $O $ 
My Approach:
Join $OM$ and $ON $as follows:

Since,$MB$ is the tangent to the circle
$\implies \angle OMB =90 ° $
Similarly, $ \angle ONB =90 ° $
Also, as $ABC$ is a quadrant
$\implies \angle MBN =90 ° $
$\implies  OMBN $ is a square
$\implies \angle NOB =45 ° $
So,we need to prove now that $\angle PON =135°$ 
So,we redraw the given figure as follows:

Here,$\angle MPN=45°$ as $ \angle MON =90°$
So, in $\triangle MPN$ ,if we can prove that $MP=NP$ ,then $\angle PMN= 67.5°$, which will give  $\angle PON =135°$ 
$ \implies $ we need to prove $∆MPO \cong ∆NPO$
So, how to prove $\triangle MPO \cong \triangle NPO $ ?
 A: You've made a big deal out of it.
Complete the circle from the quadrant for a better understanding.
The smaller circle touches the larger circle at the point $P$. Draw a tangent to the smaller circle at the point $P$. Since the larger circle touches the smaller circle at $P$ as well, the same tangent is also a tangent to the larger circle. 
Now, recall:

the perpendicular drawn from a tangential line at the point of tangency passes through the centre of the circle

Therefore, if we draw a perpendicular from the tangent at $P$, it will pass through the center $O$ as well as the center $B$ as it is a tangent to both the circles. Since the perpendicular at $P$ passes through $O$ and $B$ both, it is obvious that the line $PB$ passes through $O$.
A: 
Here is a proof with coordinates to assure rigor.
Since M and N are the tangent points of the inscribed circle, we have OM$\perp$AB, ON$\perp$BC. Then, the center has the coordinates O($r$,$r$) and $\angle$ABO = 45$^\circ$.
Let the quadrant circle be $x^2+y^2=R^2$ and the coordinates of P($a$,$b$). Then, the slope at the point P is $y' = -\frac ab$. Since OP is perpendicular to the tangent line at the point P, we have 
$$-\frac1{y'} = \frac ba = \frac {b-r}{a-r} $$
which yields $a=b$. Therefore, P is midpoint of the arc APB and $\angle$ABP = 45$^\circ$. From $\angle$ABO = 45$^\circ$ obtained earlier, we have OB||PB, which means PB passes through O.
