Question

The locus of point of intersection of lines x cosp+ y sinp =a and x sinp- y cosp =b (p is a variable)?

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{equation}\mathrm{is}, \\ \mathrm{xcosp}+\mathrm{ysinp}=\mathrm{a} \\ \mathrm{xsinp}-\mathrm{ycosp}=\mathrm{b} \\ \mathrm{Squaring}\mathrm{and}\mathrm{adding}\mathrm{both}\mathrm{the}\mathrm{equation}\mathrm{we}\mathrm{get}, \\ {\mathrm{x}}^2{\cos }^2\mathrm{p}+{\mathrm{y}}^2{\sin }^2\mathrm{p}+2\mathrm{xycospsinp}+{\mathrm{x}}^2{\sin }^2\mathrm{p}+{\mathrm{y}}^2{\cos }^2\mathrm{p}-2\mathrm{xycospsinp}={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{x}}^2\left({\cos }^2\mathrm{p}+{\sin }^2\mathrm{p}\right)+{\mathrm{y}}^2\left({\cos }^2\mathrm{p}+{\sin }^2\mathrm{p}\right)={\mathrm{a}}^2+{\mathrm{b}}^2 \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\left(\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\right)}^2 \\ \mathrm{Now}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{center}\mathrm{at}\left(0,0\right)\mathrm{and}\mathrm{radius}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \end{gathered}$$