I am new to using Tikz, so forgive my code as there is probably a much easier way to do this. I am trying to draw a point where segment OP and the circle intersect. Is there an easy way to do this? I am using tkz-euclid package. Here is my code:
\documentclass[border=10pt,tikz]{exam} \usepackage{tkz-euclide} \usepackage{amsfonts,amsmath,amssymb,tikz} \usepackage{multicol} \usepackage{pgfplots} \usepackage{tcolorbox} \usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings} \pgfplotsset{every x tick label/.append style={font=\footnotesize, yshift=0.5ex}} \pgfplotsset{every y tick label/.append style={font=\footnotesize, xshift=0.5ex}} \usepgfplotslibrary{polar} \pagestyle{empty} \usepackage{pgfplots} \pgfplotsset{compat=1.16} \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[scale=1.5] \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} \tkzDefPoint(4,3){C} \tkzDrawSegment[color=blue,thick](A,B) \tkzDrawSegment[color=blue,thick](B,C) \tkzDrawSegment[color=blue,thick](A,C) \tkzLabelSegment[right](B,C){$y$} \tkzLabelSegment[swap](A,B){$x$} \tkzLabelSegment[above](A,C){$r$} \tkzMarkRightAngle(A,B,C) \node at (5,0) [anchor=north]{$x$}; \node at (0,5)[anchor=west]{$y$}; \tkzLabelAngle[pos=.5](B,A,C){\footnotesize{\color{red}{$\theta$}}} \tkzMarkAngle[size=0.35,mark=](B,A,C) \draw[-latex,thin](-3,0)--(5,0); \draw[-latex,thin](0,-1.5)--(0,5); \node at (0,0) [anchor=east]{$O$}; \node at (4,3) [anchor=south]{$P(x,y)$}; \tkzDrawPoint(C) \draw[color=red,thick](0,0) circle (1.5); \end{tikzpicture} \end{document}
Thank you for your help and time.