Differentiation of a circle

Question

As a discus thrower is spinning counterclockwise to throw a discus, the discus travels along the path given by the circle $x^2+y^2=4$. If the discus is released at the point $(\sqrt2,\sqrt2)$ and travels along a path tangent to the circle at the point of release for a total of $200$ feet, where does the discus land? Give the $(x,y)$-coordinates.

I have no idea where to begin! :(

Answer 1

Hints: Calculus is actually unnecessary to solve this question. [If you must, then use implicit differentiation to find the slope of the tangent line at the given point.] Here's an outline to solve the problem via pre-calculus knowledge:

• Find the slope $m_r$ of the radius connecting the center to the point of tangency.
• Since radii are perpendicular to tangents in a circle, the slope of the tangent line is $m$, the negative reciprocal of $m_r$.
• Using the slope-point formula, the equation of the tangent line is: $$ y - \sqrt 2 = m(x - \sqrt 2) \tag 1 $$
• Recall that the equation for the circle centred at the point of tangency with radius $200$ is given by: $$ (x - \sqrt 2)^2 + (y - \sqrt 2)^2 = 200^2 \tag 2 $$
• Solve the system of equations consisting of $(1)$ and $(2)$. You will get two intersection points; be sure to use the point that is in the counterclockwise direction.