Show that the angle between the tangent at any point `P` and the line joining `P` to the origin `O` is the same at all points on the curve `log(x^2+y^2)=ktan^(-1)(y/x)` .

Explanation

Given, `log(x^2+y^2)=ktan^(-1)(y/x)`


Differentiate both sides


`((1)/(x^2+y^2))(2x+2y(dy)/(dx))=k((x^2)/(x^2+y^2)) ((x(dy)/(dx)-y)/(x^2))`


`(2x+2y(dy)/(dx))=k(x(dy)/(dx)-y)`


`(dy)/(dx)=(2x+ky)/(kx-2y)`


`m_1=((dy)/(dx))_(P(x_1,y_1))=(2x_1+ky_1)/(kx_2-2y_2)=` slope of tangent


`m_2=(y_1-0)/(x_1-0)=(y_1)/(x_1)=` slope of line `OP`


`tan theta = |(m_1-m_2)/(1+m_1*m_2)|= |((2x_1+ky_1)/(kx_2-2y_2)-(y_1)/(x_1))/(1+((2x_1+ky_1)/(kx_2-2y_2))*(y_1)/(x_1))|`


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