This is a question from the beginning of Edwin Moise's book, Calculus, ed. 1967, section 2.4, number 6.
I've tried solving it by different methods, but none of them worked or, if they did, they wouldn't correspond to the answer on the back of the book, namely, $P_1P_2$: $a$$|y1|$/$|x1|$.
I have tried to calculate the slope of the segment from the center to the point $P_1$, and I got $m =y1/x1$. I imagine that the tangent line to $P_1$ would have slope $m2 = -x1/y1$, since they're perpendicular to each other, but the book's answer makes me think I'm wrong. I know that the line segment from the center to $P_1$ has length $a$, and that the point $P_2$ has its y-coordinate equal to zero, since it's where the point $P_1$ crosses the x-axis. I thought I could solve it using the distance formula, by dropping a perpendicular line from $P_1$ to the x-axis, but then I remembered that I don't know what the base of the triangle would be.
I don't know how to get that answer. I can't understand what that $a$ is doing there, and why it isn't $-x1/y1$. What am I doing wrong? Thank you.
Given that $P_1$$\leftrightarrow$ $(x1, y1)$ lies on the circle $$ x^2 + y^2 = a^2$$ with $$x1 \neq 0.$$ Let $P_2$ be the point where the tangent at $P_1$ crosses the x-axis. Find the distance $P_1P_2$. [Warning: Geometric distances are never negative.]
Hint:- Let $P_2$ be (h, 0).
1) $x_1(h - x_1) = y_1^2$. Therefore, $(h - x_1)^2 = \dfrac {y_1^4}{x_1^2}$.
2) Use Pythagoras theorem. Then, $d^2 = ... = \dfrac {y_1^2}{x_1^2}(x_1^2 + y_1^2) = \dfrac {y_2^2}{x_2^2}a^2$.
The last equality shows the use of "a".