Explanation

(b)
Since, the hypocenter of a right angled triangle inscribed in a circle is a diameter of the circle. If the coordinates of the end $C$ of the hypotenuse $BC$ are $(a,\ b)$, the coordinates of $B$
[image:11_image12.emf]
are$( - a,\ - b)$. Equation of $BC$ is$\frac{y}{x} = \frac{b}{a}$. If $A$ is the vertex of the isosceles triangle, then $OA$ is perpendicular to $BC$ and the equation of $AO$ is $\frac{y}{x} = - \frac{a}{b}$ which meets the circle $x^{2} + y^{2} = r^{2}$ at points for which
$\left( \frac{a^{2}}{b^{2}} + 1 \right)x^{2} = r^{2} = a^{2} + b^{2}$
$\lbrack\because(a,\ b)\ lies\ on\ x^{2} + y^{2} = r^{2}\rbrack$
$\Rightarrow x^{2} = b^{2}\ \ \Rightarrow \ \ x = \pm b$
$\Rightarrow y = \pm a$
$\therefore$ Coordinates of A are $( - b,\ a)$ or $(b,\ - a)$