Question

The radius of the circle passing through the point of intersection of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=0$

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

Answer 1

The correct option is B $\frac{\sqrt{2}ab}{\sqrt{a^2+b^2}}$ unit

Two curve are symmetrical about both axes and intersect in four points, so, the circle through their point of intersection will have centre at origin.
Solving $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=0,$ we get
$x^2=y^2=\frac{a^2{b}^2}{a^2+b^2}$
$\therefore$ radius of the circle $=\sqrt{\frac{2a^2{b}^2}{a^2+b^2}}=\frac{\sqrt{2}ab}{\sqrt{a^2+b^2}}$ unit