Question

Let the probability that distance from any point selected at random on the circle $x^2+y^2=16$ to the line $4x+3y=25$ is less than $2$ , is given by $\frac{2}{a\pi }{\left(cos\right)}^{-1}\left(\frac{b}{c}\right)$ , where $a,b,c\in N\&b$ and $c$ are co-prime, then $\frac{12}{a}cos\left(\frac{c\pi }{b}\right)$ is equal to

Answer 1

Answer: 3

Circle $x^2+y^2=16$ , Centre $\left(0,0\right)$ , Radius $=4$
Distance from centre of circle to the given line is $\left|\frac{-25}{\sqrt{4^2+3^2}}\right|=5$
Line parallel to $4x+3y=25$ & at a distance $3$ from centre of circle is $4x+3y+\lambda =0$

$OP=3\Rightarrow \left|\frac{0+0+\lambda }{\sqrt{25}}\right|=3\Rightarrow \lambda =15,-15$
For $4x+3y=-15,d>2$

From the figure, $cos\theta =\frac{3}{5}$
$\therefore$ required probability $=\frac{R\left(2\theta \right)}{2\pi R}$
$=\frac{\theta }{\pi }=\frac{1}{\pi }{\left(cos\right)}^{-1}\left(\frac{3}{5}\right)$
Comparing with $\frac{2}{a\pi }{\left(cos\right)}^{-1}\left(\frac{b}{c}\right),a=2,b=3,c=5$
Now, $\frac{12}{a}cos\left(\frac{c\pi }{b}\right)=\frac{12}{2}cos\left(\frac{5\pi }{3}\right)=\frac{12}{4}=3$