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Suppose I have f(x) = -3x + 4, along which (3,-5) sits as the center of a circle with a radius of 5. I want to find the point (x2, y2) along the outside of the circle that is the point of tangency for the tangent line that is perpendicular to f(x). Let's call that line g(x). We can solve for (x2,y2) by using the substitution method of solving systems of equations. However, we do not know the equation for g(x). We know that the slope of g(x) is simply -1/slope of f(x). However, the problem is that the y-intercept of g(x) is unknown. We do not technically know any points on this line as of yet.

I'm aware that there are an infinite number of tangent lines for a circle, and because we have not isolated a single point on our line as of yet, we have not isolated any one or set of tangent lines. There must be a way to isolate the tangent line that we want and then go forward and solve for our point of tangency (x2, y2), but I am unaware of an methods.

Any help would be appreciated.

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Hint:

The equation of the circle is $$ (x-3)^2+(y+5)^2=25 $$ and the points of tangency are the points of intersection of the circle with the line $y=-3x+4$ since , in these points, the tangent is orthogonal to the diameter.

If you want to use the equation of the tangent line: $y=\frac{1}{3}x+q$, note that this line is tangent to the circle if the system: $$ \begin{cases} (x-3)^2+(y+5)^2=25\\ y=\frac{1}{3}x+q \end{cases} $$ has only one real (double) solution, that is if the discriminant of the equation that we find substituting the second equation in the first is null.

This discriminant $\Delta$ is a polynomial of second degree in $q$. Solve the equation $\Delta=0$ and find the values of $q$ for the tangent lines. And after this you can easily find the points of tangency.

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The point $P$ you're looking for will be at distance $R$ (in this case $R=5$) from the center and $P \in r$ where $r \equiv y = -3x + 4$. So

$$ P = (3, -5) + \lambda(1, -3) $$

and

$$ ||\lambda(1, -3)|| = \lambda||(1, -3)|| = \lambda \sqrt{10} = 5 $$

That will give you a positive $\lambda$, but note that $-\lambda$ yields a solution too (the opposite point in the circumference).

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    $\begingroup$ I'm sorry, I do not understand lambda calculus. I am currently a second year calc student. $\endgroup$ – Brady Burnsides Sep 12 '16 at 1:38
  • $\begingroup$ You might find it easier if you consider $\lambda \in \Bbb R$ just a scale factor $\endgroup$ – cronos2 Sep 12 '16 at 7:41

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