Internal Tangency Theorem

this is just a question I thought of while doing the circles chapter of my school geometry textbook.

If there are two points $P$ and $R$ bound inside a circle $O$, is it always possible to draw a circle $Q$ that is internally tangent to $O$ and passes through points $P$ and $R$?

Some thoughts I have regarding this problem and how to prove or disprove this fact, is the fact that three points define a circle, so no matter what there will always be a unique circle defined by $P$, $R$ and the tangency point of circles $O$ and $Q$. I am stuck on how to prove that circle $Q$ is always internally tangent to $O$?

Since points $P$ and $R$ are on circle $Q$, then $PR$ is a chord in $Q$. Therefore, the center $C$ of $Q$, lies on the perpendicular bisector of $PR$. The parametric equation of this line is

$ C = M + t u $

where $M = \dfrac{1}{2} (P + R) $ and $ u \perp (R - P) $ and $ \| u \| = 1 $

Now the condition we have on $C$ is its distance from point $P$ (and equivalently from point $R$) is equal to $r - \| C \|$ where $\| C \|$ is the distance from $C$ to the origin (assumed here to be the center of circle $O$) and $r$ is the radius of circle $O$. Squaring these distances and equating

$ r_1^2 = (PC)^T (PC) = (r - \sqrt{ C^T C} )^2 $

where $r_1$ is the radius of $Q$. Plugging in expressions for $PC$ and $C$,

$ ( (M - P) + t u )^T ( (M - P) + t u ) = r^2 - 2 r \sqrt{C^T C} + C^T C $

and

$C^T C = M^T M + 2 t M^T u + t^2 $

Plugging this into the first equation

$ (M-P)^T (M - P) + 2 t (M - P)^T u + t^2 = r^2 - 2 r \sqrt{C^T C} + M^T M + 2 t M^T u + t^2 $

Cancelling equal terms, reduces this to

$ -2 M^T P + P^T P - 2 t P^T u = r^2 - 2 r \sqrt{C^T C} $

Hence,

$ C^T C = \left( \dfrac{r^2 + 2 M^T P - P^T P + 2 t P^T u}{2 r} \right)^2$

Define $a = \dfrac{ r^2 + 2 M^T P - P^T P }{2 r} $ and $ b = \dfrac{ 2 P^T u}{2 r} $

and plug in $C^T C $ on the left hand side, we get

$ M^T M + 2 t M^T u + t^2 = (a + b t)^2 $

Re-arranging,

$ (1 - b^2) t^2 + ( 2 M^T u - 2 a b ) t + (M^T M - a^2) = 0 $

Solving this quadratic equation results in two values of $t$, which we use to find the center $C$ and the radius $r_1$.

Shown below is an example with $r = 10$, $P = (1, 3), Q = (5, 6) $