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?
Best Answer
Here is a geometric version, not using a single formula. Start with the points $A$ and $B$ and a line $\ell$ through $A$ (see the figure below).
Construct the perpendicular line to $\ell$ through $A$ (the $\color{red}{\text{red}}$ line). Construct the perpendicular bisectors between $A$ and $B$ (the $\color{green}{\text{green}}$ line, the green dot is the midpoint of $A$ and $B$). The intersection of both constructed lines is the circle's center. The readius is the distance of the center to $A$.
You can translate every step into a formula to solve it numerically if necessary.