Question

A circle passing through the points $A(1,0)$ and $B(5,0)$ and touches the Y-axis at $C(0,\lambda )$. If $\angle$ACB is maximum, then?

• A. $|\lambda |=\sqrt{5}$
• B. $|\lambda |=2\sqrt{5}$
• C. $|\lambda |=3\sqrt{5}$
• D. $|\lambda |=4\sqrt{5}$

Answer 1

The correct option is A $|\lambda |=\sqrt{5}$

Eq. of AC$=0-\lambda =m(1-0)⟹m=-\lambda$

Eq. of BC$=0-\lambda =m(5-0)⟹m=\frac{-\lambda }{5}$

Slope of Line AC $\tan \theta =-\frac{\lambda }{1}$

Slope of Line BC $\tan {\theta }_1=-\frac{\lambda }{5}$

$\angle ACB=\pi +{\tan }^{-1}\frac{\lambda }{1}-\pi -{\tan }^{-1}\frac{\lambda }{5}$

$\angle ACB=-{\tan }^{-1}\frac{\lambda }{5}+{\tan }^{-1}\frac{\lambda }{1}$

$\frac{d\angle ACB}{d\lambda }=-\frac{5}{{\lambda }^2+25}+\frac{1}{{\lambda }^2+1}$

Equating to zero, we get -

$5{\lambda }^2+5={\lambda }^2+25$

$4{\lambda }^2=20$

$|\lambda |=\sqrt{5}$