There is a circle of radius 187.5m and centre $(25.23,188.6)$.
It takes the form $(x-25.23)^2 +(y-188.6)^2=187.5^2$
There is another circle whose centre is unknown and has a radius of 1750m. This circle intersects with the y axis at $(0,2.05)$ and a point on the original circle. The point on the original circle is the point where the tangent line from point $(0,2.05)$ touches the larger circle. I want to obtain the centre of this circle.
I've created this graphically in CAD software, and the resultant centre should be close to $(87.5028,1749.861)$. I've tried doing this mathematically but i fear I haven't dealt with quadratics in a long time. Sometimes I get close to the point provided by CAD but when verifying the circle equation by plugging the point $(0,2.05)$ back in, it doesn't converge.
I've calculated the tangent point in this case to be $(16.752,1.29$) which I believe is correct.
However my problem lies when trying to get the centre point of the larger circle. If both points are on the circle then the following is true:
$(0-h)^2 + (2.05-k)^2 = 1750^2$
and
$(16.752-h)^2 + (1.29-k)^2 = 1750^2$
I solve for h in terms of k in one equation. Reinput h back into the other equation and solve for k. Once I solve for the values of k, I then solve for h. But I've used online calculators and still can't seem to reconcile the numbers.
Could someone possibly explain if my work is flawed in some way? Or possibly identify that it is possible to get similar numbers to the graphical solution? I would provide the graphic but the scales of the circles are so large it's difficult to see the area I'm concentrating on.
Thanks
Best Answer
I couldn’t tell you why these calculators that you haven’t identified don’t give you solutions that match your CAD approximation, but keep in mind that there are two solutions to that equation, corresponding to the fact that there are two circles with the given radius that pass through the two points. When I give your equations to Mathematica to solve, one of the solutions jibes with your CAD numbers. The usual way to go about solving this system by hand is to subtract one equation from the other to get the equation of a line (in fact, the perpendicular bisector of the tangent segment), solve that for one of the variables and back-substitute to get a quadratic equation in one variable.
However, once you have the point of tangency (call it $B$), there’s no need to solve any equations at all in order to find the center of the large circle. It can be done with simple vector computations and an application or two of the Pythagorean theorem. Call the known $y$-axis intersection of the large circle $A$. It’s center lies somewhere on the perpendicular bisector of $AB$. This line has for a direction vector $(A_y-B_y,B_x-A_x)$. Using your values and normalizing, this gives us the unit vector $\mathbf u\approx(0.0453,0.9990)$. Half of the distance from $A$ to $B$ is $8.385$, so the distance of the large circle’s center from the midpoint of $AB$ is $\sqrt{1750^2-8.385^2}\approx1749.98$. The center of the large circle is therefore at $$\frac12(A+B)\pm1749.98\mathbf u.$$ Taking the positive direction along $\mathbf u$ again produces values that are pretty close to your estimate.
Since the distance between $A$ and $B$ is so small relative to the large circle’s radius and the tangent line is almost horizontal, you have to take extra care to manage roundoff error in your calculations. Also, there are two tangents to a circle through any exterior point, so as you’ve described the problem here, there are four solutions. No doubt you have some other critieria with which to select the correct one, but you haven’t mentioned them.