# Find the angle to intersection of the following curves :

Question:

Find the angle to intersection of the following curves :

$x^{2}+y^{2}-4 x-1=0$ and $x^{2}+y^{2}-2 y-9=0$

Solution:

Given:

Curves $x^{2}+y^{2}-4 x-1=0 \ldots(1)$

$\& x^{2}+y^{2}-2 y-9=0$

First curve is $x^{2}+y^{2}-4 x-1=0$

$\Rightarrow x^{2}-4 x+4+y^{2}-4-1=0$

$\Rightarrow(x-2)^{2}+y^{2}-5=0$

Now, Subtracting $(2)$ from $(1)$, we get

$\Rightarrow x^{2}+y^{2}-4 x-1-\left(x^{2}+y^{2}-2 y-9\right)=0$

$\Rightarrow x^{2}+y^{2}-4 x-1-x^{2}-y^{2}+2 y+9=0$

$\Rightarrow-4 x-1+2 y+9=0$

$\Rightarrow-4 x+2 y+8=0$

$\Rightarrow 2 y=4 x-8$

$\Rightarrow y=2 x-4$

Substituting $y=2 x-4$ in (3), we get,

$\Rightarrow(x-2)^{2}+(2 x-4)^{2}-5=0$

$\Rightarrow(x-2)^{2}+4(x-2)^{2}-5=0$

$\Rightarrow(x-2)^{2}(1+4)-5=0$

$\Rightarrow 5(x-2)^{2}-5=0$

$\Rightarrow(x-2)^{2}-1=0$

$\Rightarrow(x-2)^{2}=1$

$\Rightarrow(x-2)=\pm 1$

$\Rightarrow x=1+2$ or $x=-1+2$

$\Rightarrow x=3$ or $x=1$

So, when $x=3$

$y=2 \times 1-4$

$\Rightarrow y=2-4=-2$

The point of intersection of two curves are $(3,2) \&(1,-2)$

Now ,Differentiating curves (1) \& (2) w.r.t $x$, we get

$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-1=0$

$\Rightarrow 2 \mathrm{x}+2 \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}-4-0=0$

$\Rightarrow \mathrm{x}+\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}-2=0$

$\Rightarrow \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=2-\mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2-\mathrm{x}}{\mathrm{y}} \ldots$ (3)

$\Rightarrow x^{2}+y^{2}-2 y-9=0$

$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}-2 \frac{d y}{d x}-0=0$

$\Rightarrow x+y \cdot \frac{d y}{d x}-\frac{d y}{d x}=0$

$\Rightarrow x+(y-1) \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{-x}{y-1} \ldots(4)$

At $(3,2)$ in equation $(3)$, we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2-3}{2}$

$\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{2}$

At $(3,2)$ in equation(4), we get

$\Rightarrow \frac{d y}{d x}=\frac{-3}{2-1}$

$\Rightarrow \frac{d y}{d x}=-3$

$\Rightarrow m_{2}=\frac{d y}{d x}=-3$

when $m_{1}=\frac{-1}{2} \& m_{2}=0$

$\Rightarrow \tan \theta=\left|\frac{\frac{-1}{2}-3}{1+\frac{-1}{2} \times 3}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-7}{2}}{1+\frac{-3}{2}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-7}{2}}{\frac{-1}{2}}\right|$

$\Rightarrow \tan \theta=7$

$\Rightarrow \theta=\tan ^{-1}(7)$

$\Rightarrow \theta \cong 81.86$