# Calculating vertical displacement to intercept the perimeter of the circle after a horizontal shift

I can't come upon a solution for the question below. The background of the actual work is not interesting for the question, so l won't explain why I need the solution. I would be very thankful if someone could help, even though the question may be pretty simple for the Mathematica Stack Exchange.

I have a circle with a radius of 184 cm. I moved a point placed on the top of the circle horizontally 17 cm to the left. Now, I want to move this point vertically downward such that it intercepts the perimeter of the circle. How many centimeters should I move the point downward to intersect the perimeter of the circle again? How can this problem be set up with Mathematica?

• Might this be a question for the Mathematics page, rather than the Mathematica page? Mar 27 at 22:47

Assuming the circle with radius: r is at the origin. Then the top point has coordinates:

r=184;
top= {0,r}


The point to the left with a displacement: dis of has coordinates:

dis= -17;
l= {dis,r}


The point left down on the circle: ld makes an angle phi with the vertical axis:

phi= ArcTan[r,dis]


And the point on the circle:

ld= r {Sin[phi],Cos[phi]}


The distance you are searching is then:

dely= r-ld[[2]]

184 - 33856/Sqrt[34145]

• Thanks for your detailed explanation! Mar 28 at 19:40
sol = First@
Solve[{x^2 + y^2 == 184^2, x == -17, y > 0}, {x, y}, Reals]


{{x -> -17, y -> Sqrt[33567]}}

x = 184 - y /. sol


184 - Sqrt[33567]

Geometric solution

d1 = Disk[{0, 0}, 184];
c1 = d1 /. Disk :> Circle;
r1 = Rectangle[{0, 0}, {-17, 184}];

sol = RegionIntersection[c1, RegionBoundary@r1]


Point[{{0, 184}, {-17, Sqrt[33567]}}]

GraphicsRow[{
Graphics[{d1, Red, r1, Yellow, AbsolutePointSize@Large,
Point@RegionCentroid@d1}]
, Graphics[{c1, FaceForm[None], EdgeForm[Thick], r1,
AbsolutePointSize@Large, Red, sol}
, PlotRange -> {{-20, 10}, {180, 186}}]
}]


• That’s an awesome solution, thanks Mar 28 at 19:44