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Given the point $(x_1,y_1)$,and the two lines $f(x)=m(x-x_1)+y_1$ and $g(x)=m_1x+b_1$, I'm trying to find the the equation of the circle that goes through $(x_1,y_1)$ and is tangent to both of the given lines. Finding either the radius of the circle or the center of the circle would lead to an easy solution. Apparently there is a way to find the radius of the circle per this answer, but it went completely over my head. I personally couldn't think up any ways to find the radius of the circle so I started thinking about finding its center. An important fact for this strategy is that the line that is equidistant from any two points on a circle always goes through the center of that circle, and the second and more obvious fact is that the perpendicular through the point in which a tangent is tangent also goes through the center of the circle. Suppose that we know the center of the circle and that it is located at $(c_1,c_2)$, then the perpendicular to $g$ through the center is $g^\perp(x)=-\frac{1}{m_1}(x-c_1)+c_2$ and the intersection between $g^\perp$ and and $g$ is $(\frac{c_1+m_1(c_2-b_1)}{m_1^2+1},\frac{m_1^2c_2+m_1c_1+b_1}{m_1^2+1})$. The line of points equidistant from $(x_1,y_1)$ and $(\frac{c_1+m_1(c_2-b_1)}{m_1^2+1},\frac{m_1^2c_2+m_1c_1+b_1}{m_1^2+1})$ is a seriously complicated one that I have personally failed to simplify, but theoretically could be used in in conjunction with $f$ to find the center as $f$ should only intersect it once at the center. If anyone had any explanations of the answer that I linked to or explanations for how to more easily find the center of the circle without explicitly knowing the radius that would be greatly appreciated.

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  • $\begingroup$ use the distance formula wiki $\endgroup$ – Lozenges Jun 20 '18 at 5:38
  • $\begingroup$ Are your two lines parallel, that is, is the $m$ in the first line’s equation meant to also be $m_1$ instead? $\endgroup$ – amd Jun 20 '18 at 6:48
  • $\begingroup$ See this cut the knot page. $\endgroup$ – Jan-Magnus Økland Jun 20 '18 at 8:53
  • $\begingroup$ Unless I'm missing something, you haven't made use of the fact that f is tangent to the circle at ($x_1$, $y_1$). $\endgroup$ – Steve B Jun 20 '18 at 10:25
  • $\begingroup$ The two lines are not meant to be parallel. Stevw, you are correct, I have yet to use the fact that $f$ is tangent at $(x_1,y_1)$. $\endgroup$ – Aaron Quitta Jun 20 '18 at 16:29
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Solving this problem is a matter of computing the intersections of pairs of lines. The point $(x_1,y_1)$ lies on one of the lines, so the condition that the circle is tangent to the line at that point means that its center lies on the perpendicular line through that point. The condition that the circle is tangent to two lines means that its center lies on one of the angle bisectors of those lines. Assuming a non-degenerate configuration in which the given tangent point is not the intersection of the two lines, there will be two solutions. If the lines happen to be parallel, there’s only one solution, but it can be found the same way.

I find it a bit easier to work with the equations in a different form:$$mx-y-(mx_1-y_1)=0 \\ m_1x-y+b_1=0.$$ The equations of the angle bisectors, derived from the point-line distance formula, are then $${mx-y-(mx_1-y_1) \over \sqrt{1+m^2}} = \pm{m_1x-y+b_1 \over \sqrt{1+m_1^2}}.$$ An equation of the perpendicular line through $(x_1,y_1)$ is $x+my=x_1+my_1$. Computing the intersections of these lines should be somewhat tedious, but straightforward.

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