Question

The circle with equation $x^2+y^2=1$ intersects the line $y=7x+5$ at two distinct point A and B. Let C be the point at which the positive x-axis intersects the circle. The $\angle ACB$ is

• A. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• C. ${\tan }^{-1}\left(1\right)$
• D. ${\tan }^{-1}\left(\frac{3}{2}\right)$

Answer 1

Answer: (C)

${\tan }^{-1}\left(1\right)$

Point of intersection (h,k) of line with the circle
$k^2+(7k+5{)}^2=1$
$\Rightarrow k=\frac{-3}{5},\frac{-4}{5}$
$\Rightarrow h=\frac{4}{5},\frac{-3}{5}$
So, A ($\frac{4}{5},\frac{-3}{5}$) ; B ($\frac{-3}{5},\frac{-4}{5}$)
C is the point where circle intersects positive X axis i.e $C(1,0)$
Slope of CA is $3$ and slope of CB is $\frac{1}{2}$ . hence the angle between them is ${\tan }^{-1}1$