Question

A circle passes through the three points of an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, whose eccentric angles are $\frac{\pi }{2},\frac{\pi }{4},\frac{\pi }{4}$. Find the centre of circle.

• A. $(\frac{5}{12},-\frac{5}{8})$
• B. $(-\frac{5}{8},\frac{5}{8})$
• C. $(\frac{5}{12},\frac{5}{8})$
• D. $(\frac{5}{8},-\frac{5}{8})$

Answer 1

Answer: (A)

$(\frac{5}{12},-\frac{5}{8})$

If a circle passes through three points on ellipse whose eccentric angles are $\alpha ,\beta ,\gamma$ then its center $(h,k)$ is given by,

$h=\left\{\left(\frac{a^2-b^2}{4a}\right)(\cos \alpha +\cos \beta +\cos \gamma +\cos (\alpha +\beta +\gamma ))\right\}k=\left\{\left(\frac{b^2-a^2}{4b}\right)(\sin \alpha +\sin \beta +\sin \gamma -\sin (\alpha +\beta +\gamma ))\right\}\Rightarrow h=\left\{\left(\frac{9-4}{12}\right)(\cos \frac{\pi }{2}+\cos \frac{\pi }{4}+\cos \frac{\pi }{4}+\cos (\frac{\pi }{2}+\frac{\pi }{4}+\frac{\pi }{4}))\right\}\Rightarrow h=\frac{5}{12}(0+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)=\frac{5(\sqrt{2}-1)}{12}\Rightarrow k=\left\{\left(\frac{4-9}{8}\right)(\sin \frac{\pi }{2}+\sin \frac{\pi }{4}+\sin \frac{\pi }{4}-\sin (\frac{\pi }{2}+\frac{\pi }{4}+\frac{\pi }{4}))\right\}\Rightarrow k=\frac{-5}{8}(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0)=\frac{-5(\sqrt{2}+1)}{8}$

So, the coordinates of centre of circle are $(\frac{5(\sqrt{2}-1)}{12},\frac{-5(\sqrt{2}+1)}{8})$.

None of the options are correct.