I have $2$ points on my graph, $(-1,1)$ and $(2,1)$ and want to make an arc connecting the two with $5$ height (so it would be $6$ at one moment, because $1+5=6$). Is this possible with trigonometry? If I could have any explanation on the way to do it, that would be great!
Make an arc with 5 units height from the points (-1,0) to (2,1) with steps
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trigonometry
1 Answers
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You can do it with trigonometry, but simple geometry (no trig functions) will also do the job.
We'll find the center and radius of the circle to which the arc belongs. I hope you can draw the arc with that information.
You have two points on a circle, $(-1,1)$ and $(2,1).$ The center of the circle is necessarily on the perpendicular bisector of the segment joining those points. Since that segment is perfectly horizontal, the bisector is easy to find: it's just the vertical line through the midpoint of the segment, which is $\left(\frac12, 1\right).$ So it's the line with equation $x=\frac12.$
Since the line $x=\frac12$ passes through the center of the circle, the diameter of the circle is part of that line. Since the line is vertical, so is the diameter, and it contains the highest point on the circle. You have decided the $y$ coordinate of the highest point should be $6,$ and the function of the line gives us the $x$ coordinate, so the highest point is $\left(\frac12, 6\right).$
From here there are various ways to continue. One is, take the newly-discovered highest point and one of the points you started with; since these are two points on the circle, the center of the circle is on the perpendicular bisector of those points as well as on the line $x=\frac12.$ So we could take, for example, the two points $(2,1)$ and $\left(\frac12, 6\right),$ find the formula for the perpendicular bisector using one of the procedures shown in the answers to this earlier question and several others (here, here, or here, for example), and then find the intersection of those two lines. Conveniently, since one of the intersecting lines is $x=\frac12,$ for this intersection you just have to plug $x=\frac12$ into the equation of the other line.
The intersection of those lines is the center of the circle. The radius is the distance from the center to the point $\left(\frac12, 6\right).$
The procedure above requires only the four basic operations (multiplication, addition, division, and subtraction).
Alternatively, using trigonometry, you can find the slopes of the lines between $\left(\frac12, 6\right)$ and your two original points, and use the arc tangent of each slope to find the angle between the lines. That angle is half the angle made by the two radii from the center of the circle to the two points $(-1,1)$ and $(2,1),$ so it's the angle between one of those radii (say from the center to $(2,1)$) and the line $x=\frac12,$ and therefore it's also the angle at the top of the triangle with vertices at the circle's center, $\left(\frac12, 1\right),$ and $(2,1).$ Use the cotangent of that angle and the length of the opposite side (the segment from $\left(\frac12, 1\right)$ to $(2,1)$) to find the length of the segment from $\left(\frac12, 1\right)$ to the center of the circle, that is, the height of the center above the point $\left(\frac12, 1\right).$
I think I prefer the way without trigonometry.