Question

The acute angle between the lines joining the origin to the points of intersection of the line $\sqrt{3}x+y=2$ and the circle $x^2+y^2=4$ is

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{3}$

Explanation for the correct option:

Step 1: Finding the values of $a,h,b$

The given equation of the line is

$$\begin{gathered} \sqrt{3}x+y=2 \\ \Rightarrow \frac{\sqrt{3}x+y}{2}=1....\left(i\right) \end{gathered}$$

And the given equation of the circle is

$$\begin{gathered} x^2+y^2=4 \\ \Rightarrow x^2+y^2-4=0 \\ \Rightarrow x^2+y^2-4\times {\left(1\right)}^2=0 \end{gathered}$$

Substituting from $\left(i\right)$

$$\begin{gathered} x^2+y^2–\frac{4{(\sqrt{3}x+y)}^2}{4}=0 \\ \Rightarrow \frac{4x^2+4y^2–4\left(x^2–2\sqrt{3}xy–y^2\right)}{4}=0 \\ \Rightarrow 4x^2+4y^2–12x^2–8\sqrt{3}xy–4y^2=0 \\ \Rightarrow -8x^2–8\sqrt{3}xy=0 \\ \Rightarrow x^2–\sqrt{3}xy=0 \end{gathered}$$

Now compare it with the general homogenous equation of a curve

$x^2+y^2+2gx+2fy+2hxy+c=0$

We have $a=1,h=-\frac{\sqrt{3}}{2},b=0$

Step 2: Finding the acute angle between the lines

So, the angle between the lines is given by,

$\begin{array}{rcl}\tan \theta & =& \frac{2\sqrt{(h^2–ab)}}{(a+b)}\\ \tan \theta & =& 2\frac{\sqrt{\left({\displaystyle \frac{3}{4}}\right)-0}}{\left(1+0\right)}\\ & =& 2\sqrt{\frac{3}{4}}\\ & =& \sqrt{3}\\ \tan \theta & =& \tan 60°\\ \theta & =& \frac{\mathrm{\pi }}{3}\end{array}$

Hence, option (C) is the correct answer.