angle of intersection between two lines Show that the angle between the tangent at any point $p$ and the line joining $p$ to the origin $O$ is same at all the points of the curve $\log(x^2+y^2)=c\tan^{-1}\left(\dfrac{y}{x}\right)$ where $c$ is any arbitrary constant .
 A: Denote $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan(y/x)$ as the polar coordinates, then the given equation
$$\log(x^2+y^2)=c\tan^{-1}\left(\dfrac{y}{x}\right) \implies 2\log r = c \theta \implies r = \mathrm{Exp}\left(\frac{c}2 \theta \right)$$
is readily recognized as the logarithmic spiral.
The requested "constant angle" property is well known and easily proven, see for example Eq.(5) of this page.
If one wants to start from scratch and deal with the equation in the given form, then read on. 

The title of the original post "angle of intersection between two lines", suggests that the asker was having difficulties determining the angle between the tangent direction $(1, y')$ and radial direction $(x,y)$.
The standard way to do this is to consider the inner product. Denote $\alpha$ as the angle between the two lines (or local angle between two local directional vectors), then the task is to show
$$\cos\alpha = \frac{ \text{inner product} }{ \text{product of norms} } =\frac{ (1, y') \bullet (x,y) }{ \lVert (1,y')\rVert\,\cdot\, \lVert(x,y) \rVert} \overset{?}{=} \text{constant}$$
Here $\displaystyle y' \equiv \frac{ \mathrm{d} y }{ \mathrm{d}x}$. Let's take the implicit derivative on both sides of the entire give equation.
\begin{gather*}
&\frac1{x^2 + y^2} \left( 2x + 2yy'\right) = c \frac1{ 1 + (y/x)^2} \frac{ y' x - y }{ x^2 } \\
&\implies 2\left( x + yy'\right) = c(y' x - y) \implies y' = \frac{ 2x + cy }{ cx - 2y} \tag{1} \label{1}
\end{gather*}
This gives the inner product as 
$$(1, y') \bullet (x,y) = x+yy' = x + y\frac{ 2x + cy }{ cx - 2y} = \frac{  x(cx - 2y) + y (2x + cy) }{ cx - 2y} = \frac{  c(x^2 + y^2)}{ cx - 2y} \tag{2} \label{2}$$
Now, also make use of Eq.\eqref{1} to obtain the product of the norms:
\begin{gather*}
\lVert(1,y')\rVert\,\cdot\, \lVert(x,y)\rVert = \sqrt{ \left( 1+(y')^2 \right) \left( x^2 + y^2\right)} \\ 
\text{with} \quad 1+(y')^2 = \frac{ (cx - 2y)^2 + (2x + cy)^2 }{ (cx - 2y)^2 } \implies 1+(y')^2 = \frac{ (c^2 + 4)( x^2 + y^2) }{ (cx - 2y)^2 } \\
\implies \lVert(1,y')\rVert\,\cdot\, \lVert (x,y) \rVert = \sqrt{(c^2 + 4)}\frac{ x^2 + y^2 }{ |cx - 2y| } \tag{3} \label{3}
\end{gather*}
Finally, putting Eq.\eqref{2} and Eq.\eqref{3} together we have
$$\cos\alpha = \frac{ \pm c }{ \sqrt{(c^2 + 4)} } \tag{4} \label{4}$$
where the sign is determined by the sign of $cx - 2y$.
We know that a cosine flipping sign means the angle flips to its complement to $\pi$. That is, Eq.\eqref{4} correctly gives the cosine of a oriented angle of a constant magnitude between two oriented directions.$~~$Q.E.D.
