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:
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\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}