How can I find the equation of a circle given two points and a tangent line through one of the points? I was wondering whether it was possible to find the equation of a circle given two points and the equation of a tangent line through one of the points so I produced the following problem:
Find the equation of the circle which passes through $(1,7)$ and $(-6,0)$ and has a tangent with equation $2x-9y+61=0$ at $(1,7)$
This seems like it should be solvable but I cannot work out how. Clearly, the line and the circle have one point of intersection so I tried finding the point of intersection between the line and the circle using the generic circle equation $(x-a)^2 + (y-b)^2 = r^2$, the equation of the line, and the discriminant of the resulting quadratic, which must be 0, but this still produces a quadratic with two unknowns.
I also feel like the fact that the perpendicular distance between centre $(a,b)$ and the line is the radius can be used somehow. Again, trying this seems to produce equations with too many unknowns.
How can I solve this problem?

Hint. The center of such circle is on the line which is orthogonal to tangent line and passes through the point of tangency. 
Therefore, in your case, the coordinate of the center is $C=(1+2t,7-9t)$ for some $t\in \mathbb{R}$. In order to find $t$, impose that $C$ has the same distance from the given points $P=(1,7)$ and $Q=(−6,0)$:
$$|CP|^2=|CQ|^2\Leftrightarrow (4+81)t^2=(7+2t)^2+(7-9t)^2.$$
