Question

The radius of the circle passing through the points of intersection of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$= 1 and x2 – y2 = 0 is –

• A. $\frac{ab}{\sqrt{a^{2} + b^{2}}}$
• B. $\frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}}$
• C. $\frac{a^{2} - b^{2}}{\sqrt{a^{2} + b^{2}}}$
• D. $\frac{a^{2} + b^{2}}{\sqrt{a^{2}–b^{2}}}$

Answer 1

Answer: (B)

$\frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}}$

Answer 2

Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.

Solving x2 – y2 = 0 and $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$= 1, we get

x2 = y2 = $\frac{a^{2}b^{2}}{a^{2} + b^{2}}$

Therefore radius of circle

=$\sqrt{\frac{2a^{2}b^{2}}{a^{2} + b^{2}}}$ =$\frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}}$