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Equation of Circle Whose Extremities of a Diameter Given

An isosceles ...

Question

An isosceles right angled triangle is inscribed in the circle $x^{2} + y^{2} = r^{2} .$ If the coordinates of an end of the hypotenuse are (a, b), the coordinates of the vertex are

(-a, -b)

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(b, -a)

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(b, a)

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(-b, -a)

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Solution

The correct option is B
(b, -a)

Since the hypotenuse 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 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 circles $x^{2} + y^{2} = r^{2}$ at points for which
$(\frac{a^{2}}{b^{2}} + 1) x^{2} = r^{2} = a^{2} + b^{2} [∵ (a, b) lies on x^{2} + y^{2} = r^{2}] \Rightarrow x^{2} = b^{2} \Rightarrow x = \pm b \Rightarrow y = \mp a ∴ Coordinates of A are (- b, a) or (b, - a) .$

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