Question

The point $(\sin \theta ,\cos \theta ),\theta$ being any real number, lie inside the circle $x^2+y^2-2x-2y+\lambda =0$, if

• A. $\lambda <1+2\sqrt{2}$
• B. $\lambda >2\sqrt{2}-1$
• C. $\lambda <-1-2\sqrt{2}$
• D. $\lambda >1+2\sqrt{2}$

Answer 1

The correct option is C $\lambda <-1-2\sqrt{2}$

$x^2+y^2-2x-2y+\lambda =0$

Radius of circle$=\sqrt{1+1-\lambda }$

$=\sqrt{2-\lambda }$

Maximum distance of $(\sin \theta ,\cos \theta )$

From center of above circle is when $\theta =\frac{5\pi }{4}$

Thus distance$=\sqrt{{(1+\frac{1}{\sqrt{2}})}^2+{(1+\frac{1}{\sqrt{2}})}^2}$\

$=(\sqrt{2}+1)$

$\therefore \sqrt{2-\lambda }>\sqrt{2+1}$ [For all points to lie in center]

$\therefore 2-\lambda >2+1+2\sqrt{2}$

$\lambda <2-3-2\sqrt{2}$

$\lambda <-1-2\sqrt{2}$