0

Does any body have the equation for calculating the co-ordinates for the centre of a known ellipse tangent to a known circle

Details of ellipse and circle1

Sketch of ellipse and circle attached the 20 degree dimension will be a variable

David K
  • 98,388
NigelMcB
  • 1
  • 1
    "a known ellipse" is a bit vague: if it is really known, so it is its center. – Intelligenti pauca Apr 30 '19 at 14:33
  • 1
    Welcome to Math.SE. I'm sorry, but it seems to me that we would need to know more about the ellipse and the circle tangent to it. Can you please edit your post and add that information? – Ertxiem - reinstate Monica Apr 30 '19 at 14:34
  • The last sentence is a little unclear. Is it actually meant to be two sentences, that is, "The 20 degree dimension will be a variable"? If so, do you mean you want a variable angle where the figure now shows a constant 20 degrees? – David K Apr 30 '19 at 15:54
  • Yes - it should have been 2 sentences, going forward I will be calculating the co-ordinates for various angles and circle radii. – NigelMcB May 01 '19 at 09:58

1 Answers1

1

Let's take the circle centered at $O=(0,0)$, with equation $$\tag{1}x^2+y^2=r^2,$$ where $r=200$, and let $(h,k)$ be the center of the ellipse, whose equation is then $$\tag{2}{(x-h)^2\over a^2}+{(y-k)^2\over b^2}=1,$$ with $a=25$ and $b=40$. We have in addition $$\tag{3}{h\over k}=\tan\alpha$$ (in your example $\alpha=20°$).

If $P=(x,y)$ is the tangency point, the tangent is perpendicular to $OP$ and its slope $y'$ satisfies then $y'= -x/y$, that is: $$ \tag{4}{b^2\over a^2}{x-h\over y-k}={x\over y}. $$ Equations $(1)-(4)$ form a system of degree 8, which can be (numerically) solved for the unknowns $x$, $y$, $h$, $k$. For $\alpha=20°$ I got four real solutions, one of them corresponding to an externally tangent ellipse in the first quadrant: $$ h= 81.53626271442751,\quad k= 224.01904067460944. $$ Another solution corresponds to an internally tangent ellipse, while the other two solutions correspond to reflections of the previous ellipses through point $O$.

Intelligenti pauca
  • 50,470
  • 4
  • 42
  • 77
  • Thank you very much for your answer Aretino. I have used CAD to check if your co-ordinates are correct and I agree.I take it that OP is the line from the circle origin and the point of tangency between the ellipse and circle. Unfortunately my maths is not up to your level and I would appreciate it if you could show me an example with my values plugged in and with working out so I can better understand your equations. – NigelMcB May 01 '19 at 13:00
  • As I wrote in my answer the solution of that system of equations is forcedly numerical: there is no hope to get expressions for $h$ and $k$ as a function of $a$, $b$, $r$, $\alpha$. To obtain the solution, I just plugged the equations into a Mathematica notebook and copied the result. You could do the same using some other software, or a package for the numerical solution of simultaneous equations. – Intelligenti pauca May 01 '19 at 13:19
  • Thank you Aretino for you quick response .I will have to see where I can go from here as i wish to make use of these calculated co-ordinates in a excel VBA program. – NigelMcB May 01 '19 at 13:44
  • As a partial correction to what I wrote above, Mathematica CAN give a symbolic expression of the solution, but it is so long and complex to be practically useless. The expression for $h$, for example, is more than 6000 lines long. – Intelligenti pauca May 01 '19 at 13:51
  • Aretino - i noticed you changed your equations in the Wolframalpha example, is there a reason for this? – NigelMcB May 02 '19 at 11:44
  • Hello again Aretino, using the equations you used in the Wolframaplha example, I have now been able to get excel to solve the equations, all i need to do now is work out the VBA code - Thanks for you help – NigelMcB May 02 '19 at 12:13
  • @NigelMcB Glad I was of help. The equations I supplied to Wolframalpha are the same as those in my answer above. I only eliminated denominators in eq. (4). – Intelligenti pauca May 02 '19 at 13:07