The following is another solution that uses "analytic geometry," at a lower level of technical sophistication. The downside is that the equations look uglier and less structured than they could be.
We will use the not very hard to prove fact that if a line has equation $ax+by+c=0$, then the perpendicular distance from $(u,v)$ to the line is $\frac{|au+bv+c|}{\sqrt{a^2+b^2}}.$ Note that in the numerator we have an absolute value. That will cause some headaches later.
(The formulas that follow would look simpler if we adjusted the equation of any line $ax+by+c=0$ by dividing each coefficient by $\sqrt{a^2+b^2}$, but we will not do that.)
Given two points on each of our lines, we can find equations for the lines. Suppose that these equations turn out to be $a_1x+b_1y+c_1=0\qquad\text{and}\qquad a_2x+b_2y+c_2=0.$
Let $(u,v)$ be the center of the circle, and let $r$ be its radius. Then from the "distance to a line" formula, we have $\frac{a_1u+b_1v+c_1}{\sqrt{a_1^2+b_1^2}}=\pm r \qquad\text{and}\qquad \frac{a_2u+b_2v+c_2}{\sqrt{a_2^2+b_2^2}}=\pm r.$
Unfortunately that gives in general $4$ possible systems of two linear equations, which correspond to the generally $4$ (in the parallel case, $3$) pieces into which the lines divide the plane. At the end of this post are some comments about how to identify which signs to choose.
But suppose for now that we have identified the two relevant equations. We can use them to "eliminate" $r$, and obtain a linear equation $ku+lv+m=0$ that links $u$ and $v$.
Since $(u,v)$ is the center of the circle, and $r$ is the radius, we have $(u-x_p)^2+(u-y_p)^2=r^2.$
Use one of our linear expressions for $r$ to substitute for the $r^2$ term. We obtain a quadratic equation in $u$ and $v$. Use the equation $ku+lv+m=0$ to eliminate one of the variables. We are left with a quadratic equation in the other variable. Solve.
Note that in general we will get two solutions, since, almost always, there are two circles that work, a small circle with $(x_p,y_p)$ on the other side of the circle from where the two given lines meet, and a big circle with $(x_p,y_p)$ on the near side of the circle from where the two given lines meet.
Sign issues: It remains to deal with how to choose the $\pm$ signs in the distance equations. One approach that works reasonably well is, in our line equations $a_ix+b_iy+c_i=0$, always to choose the coefficient of $y$ to be positive. (This can be done unless the line is vertical.) Then if $a_1x_p+b_1y_p+c_1$ is positive, use $(a_1x_p+b_1y_p+c_1)/\sqrt{a_1^2+b_1^2}=r$, and if it is negative use $-r$. Do the same with the other line. The reason this works is that that if the coefficient of $y$ is positive, then $a_ix+b_iy+c_i$ is positive for fixed $x$ and large $y$. So if $a_ix_p+b_iy_p+c_i$ is positive, then $(x_p,y_p)$ lies "above" the line, and if $a_ix_p+b_iy_p+c_i$ is negative, then $(x_p,y_p)$ lies below the line.