Express the length of XY in terms of a and b

Question

Where $a > 0$ and $b > 0$, the graph of the circle $x^2 + y^2 = b^2$ contains point $X$ in Quadrant II with x-coordinate −a. It also contains a second point $Y$ in Quadrant IV where $x = a$. Find an expression for the length of $XY$ in terms of $a$ and $b$.

Answer 1

You have that $X = (-a, \sqrt{(b^{2} - a^{2})})$ and $Y = (a,-\sqrt{(b^{2} - a^{2})})$

Then $\|X-Y\| = \|(-2a, 2\sqrt{(b^{2} - a^{2})})\| = $ $\sqrt{4a^{2} + 4(b^{2} - a^{2})} = \sqrt{4b^{2}} = 2b$

Answer 2

You just have to consider the symmetry of the situation. Since the circle has center $(0,0)$, the fact that $X$ and $Y$ have opposite $x$-coordinates and are in opposite quadrants implies that they are endpoints of a diameter of the circle, so $XY=2b$.

If you must have an algebraic solution, find the $y$-coordinates of the two points and then use the distance formula.

For $X$, we have $a^2+y^2=b^2$, so $y=\sqrt{b^2-a^2}$. Similarly, for $Y, y=-\sqrt{b^2-a^2}$.
(Note the signs for y are determined by the quadrants.)

Then, using $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, one gets

$\begin{align} XY&=\sqrt{(-a-a)^2+(\sqrt{b^2-a^2}-(-\sqrt{b^2-a^2}))^2}\\ &=\sqrt{(-2a)^2+(2\sqrt{b^2-a^2})^2}\\ &=\sqrt{4a^2+4(b^2-a^2)}\\ &=\sqrt{4b^2}\\ &=2b\\ \end{align}$