# Calculating tangent point on arc from a known arc point

I am trying to figure out how to calculate a tangent point on arc given a point on the arc (the midpoint) and the arc's radius.

I have a diagram: The two red lines that come to a point are the current lines. They meet at a point, let's say (0,-0.1). I need to basically move these lines to the blue position which is parallel to the original lines or tangent to another point (relatively far away, about (4.769,-5.238)). Finally, the two lines need to be connected with an arc that has a radius of 0.1.

All I need are the points on that arc/imaginary circle that are tangent to the other point or parallel to the red lines (not sure which is better/easier) and results in the 0.1 radius arc's midpoint curving up to the 0.1 distance as pictured.

I've included the answers in purple, but since I'm using a CAD program, I need to determine the formula or method for reaching those answers.

It is similar to this question: How to calculate the two tangent points to a circle with radius R from two lines given by three points

Except the point P still needs to be on the arc.

## 1 Answer

Are the two red lines supposed to be perpendicular to one another, and at 45 degrees to the vertical, as shown? If so, and the circle center is at $(a, b)$ and the circle radius is $r$, then the points you're looking for are at $$(a + \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r) \\ (a - \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r).$$

Since the two lines are (after the comments) not supposed to be perpendicular, let's call the angle between them $\alpha$; in your case, $\alpha = 85.702%{\circ}$. Let $$\beta = 90^{\circ} - \frac{\alpha}{2};$$

Then $\beta$ is the half-angle between one of your dotted orange lines and the vertical greenish line. The offset of the two points to either side of that vertical line is therefore $r \sin \beta$, so we get $$(a + r \sin \beta, b + r \cos \beta) \\ (a - r \sin \beta, b + r \cos \beta).$$

Applying these to your points, I get (using 3.14159 for PI, and Excel for the calculations)

right point = ( 0.073312432, -0.131990535) left point = (-0.073312432, -0.131990535)

That's in pretty close agreement with your CAD results, and it's also mathematically correct. :)

• No sorry, it kind of looks that way. I'd updated the image with the angle. – ikathegreat Jul 10 '14 at 19:31
• Brilliant thanks! – ikathegreat Jul 15 '14 at 16:09