Find the angle of intersection of the following curve:
$x^{2} + y^{2} - 4 x - 1 = 0$ and $x^{2} + y^{2} - 2 y - 9 = 0$ .

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Solution

The given curves are,
$x^{2} + y^{2} - 4 x - 1 = 0$ ----- ( 1 )
$x^{2} + y^{2} - 2 y - 9 = 0$ ----- ( 2 )
From curve ( 1 ) we get,
$\Rightarrow$ $x^{2} + y^{2} = 4 x + 1$
Substituting above in ( 2 ) we get,
$\Rightarrow$ $4 x + 1 - 2 y - 9 = 0$
$\Rightarrow$ $4 x - 2 y = 8$
$\Rightarrow$ $2 x - y = 4$
$\Rightarrow$ $y = 2 x - 4$ ----- ( 3 )
Substituting ( 3 ) in ( 1 ) we get,
$x^{2} + (2 x - 4)^{2} - 4 x - 1 = 0$
$\Rightarrow$ $x^{2} + 4 x^{2} + 16 - 16 x - 4 x - 1 = 0$
$\Rightarrow$ $5 x^{2} - 20 x + 15 = 0$
$\Rightarrow$ $x^{2} - 4 x + 3 = 0$
$\Rightarrow$ $x^{2} - 3 x - x + 3 = 0$
$\Rightarrow$ $x (x - 3) - 1 (x - 3) = 0$
$\Rightarrow$ $(x - 3) (x - 1) = 0$
$\Rightarrow$ $x = 3$ or $x = 1$
Substituting the values of $x$ in ( 3 ), we get
$y = 2 (3) - 4$ or $y = 2 (1) - 4$
$y = 2$ or $y = - 2$
$y = 2$ or $y = - 2$
$∴$ $(x, y) = (3, 2), (1, - 2)$
Differentiate ( 1 ) w.r.t. $x,$
$2 x + 2 y \frac{d y}{d x} - 4 = 0$

$\Rightarrow$ $\frac{d y}{d x} = \frac{4 - 2 x}{2 y} = \frac{2 - x}{y}$

$∴$ $m_{1} = \frac{2 - x}{y}$ ----- ( 4 )

Differentiating ( 2 ) w.r.t. $x,$
$2 x + 2 y \frac{d y}{d x} - 2 \frac{d y}{d x} = 0$

$\Rightarrow$ $\frac{d y}{d x} (2 y - 2) = - 2 x$

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

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

$∴$ $m_{2} = \frac{x}{1 - y}$ ----- ( 5 )

$C a s e I :$
$(x, y) = (3, 2)$
From ( 4 ) we get,
$m_{1} = \frac{2 - 3}{2} = \frac{- 1}{2}$

From ( 5 ) we get,
$m_{2} = \frac{3}{1 - 2} = - 3$

Now,
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$

$\Rightarrow$ $tan θ = ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 1}{2} + 3}{1 + \frac{3}{2}} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow$ $tan θ = ∣ ∣ ∣ ∣ ∣ \frac{\frac{5}{2}}{\frac{5}{2}} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow$ $tan θ = 1$

$\Rightarrow$ $θ = {tan}^{- 1} (1)$
$\Rightarrow$ $θ = \frac{π}{4}$

$C a s e I I :$
$(x, y) = (1, - 2)$
From ( 4 ) we het,
$m_{1} = \frac{2 - 1}{- 2} = \frac{- 1}{2}$

From ( 5 ) we get,
$m_{2} = \frac{1}{1 + 2} = \frac{1}{3}$

Now,
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$

$\Rightarrow$ $tan θ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 1}{2} - \frac{1}{3}}{1 - \frac{1}{6}} ∣ ∣ ∣ ∣ ∣ ∣$

$\Rightarrow$ $tan θ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 5}{6}}{\frac{5}{6}} ∣ ∣ ∣ ∣ ∣ ∣$

$\Rightarrow$ $tan θ = 1$

$\Rightarrow$ $θ = {tan}^{- 1} (1)$
$\Rightarrow$ $θ = \frac{π}{4}$

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Find the angle of intersection of the following curve:x2+y2−4x−1=0 and x2+y2−2y−9=0.

Find the angle of intersection of the following curve:
$x^{2} + y^{2} - 4 x - 1 = 0$ and $x^{2} + y^{2} - 2 y - 9 = 0$ .