This is the problem of finding the intersection of a straight line and a circle, as commented by J.M.. The more elementary method using analytical geometry, without rotate or translate the coordinate axes (which would make the computation easier$^1$), although not being a compact one, is the following (see sketch).

The equation defined by points $ S(N,J)$ and $T(M,I)$ is given by $ y-J=m(x-N),\qquad m=\frac{I-J}{M-N}\tag{1}. $ The equation of the circle centered at $R$ with radius $d=\overline{RQ}$ is $ (x-L)^{2}+(y-H)^{2}=d^{2}.\tag{2} $ You need to solve the following system $ \left\{ \begin{array}{c} y-J=m(x-N) \\ (x-L)^{2}+(y-H)^{2}=d^{2}, \end{array}\tag{3} \right. $ which is equivalent to $ \left\{ \begin{array}{c} x=\frac{y-J+mN}{m} \\ \left(\frac{y-J+mN}{m}-L\right)^{2}+(y-H)^{2}=d^{2}.\tag{4} \end{array} \right. $ Solving the quadratic equation yields (with the help of SWP): $ y=\frac{1}{ m^{2}+1 }\left( -mN+Lm+J+m^{2}H\pm \sqrt{\Delta}\right), \tag{5} $ where the discriminant is
$\begin{eqnarray*} \Delta &=&A+B, \\ \text{with } A &=&-m^{4}N^{2}+m^{4}d^{2}-m^{4}L^{2}-m^{2}J^{2}-m^{2}H^{2}+d^{2}m^{2}-2m^{3}NH, \\ B &=&2Lm^{3}H+2Jm^{2}H+2m^{4}NL-2m^{3}JL+2m^{3}JN.\tag{6} \end{eqnarray*}$
The information $\overline{RT}<\overline{RQ}<\overline{RS}$ will define the signal of the term $ \pm \sqrt{\Delta}$. The coordinates of $Q$ are $O=x,K=y$.
$^1$By making the translation $X=x-L$ and $Y=y-H$, and computing the new coordinates of the points in this $X,Y$ system, the above formulae simplify (it is equivalent to set $L=H=0$ in them). In the end they should be convert back to the original $x,y$ system.