- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
13.25. Miscellaneous review exercises on conic sections
- Two points $P$ and $Q$ are said to be symmetric with respect to a circle if $P$ and $Q$ are collinear with the center, if the center is not between them, and if the product of their distances from the center is equal to the square of the radius. Given that $Q$ describes the straight line $x + 2y - 5 = 0,$ find the locus of the point $P$ symmetric to $Q$ with respect to the circle $x^2 + y^2 = 4.$
- From the given Cartesian equation, we know that the center of the circle is at the origin. If $P$ and $Q$ are collinear with $O,$ with $O$ not being between between $P$ and $Q,$ what is the relationship between $P$ and $Q$?
- If points $P$ and $Q$ are collinear with the origin, but the origin is not between $P$ and $Q,$ then this implies that $P$ is a positive scalar multiple of $Q.$ In other words, $P = (u, v) = k(x, 2y),$ where $k > 0.$ To find $k,$ we note that the product of the lengths of $P$ and $Q$ must equal $4,$ thus $$\left(\sqrt{x^2 + 4y^2}\right)\left(k\sqrt{x^2 + ky^2}\right) = 4$$In other words, $k\left(x^2 + y^2\right) = 4,$ making $k = \frac{4}{x^2 + 4y^2}.$ Thus, the locus of $P$ is the set of points satisfying $k(x, 2y)$ where $x + 2y - 5 = 0.\quad\blacksquare$