The line PQ and the line tangent to PQ at point Q are perpendicular lines, since P is the circle center and Q is on the circle.
Perpendicular lines have the following relation about their slopes:
[tex]m_2=-\frac{1}{m_1}[/tex]
Comparing the equation of line PQ with the slope-intercept form of the linear equation (y = mx + b), we have a slope m = 2/5.
Therefore the slope of the tangent line is:
[tex]m_2=-\frac{1}{\frac{2}{5}}=-\frac{5}{2}[/tex]
Now, let's put the equation in the standard form:
[tex]\begin{gathered} y=-\frac{5}{2}x+b \\ y+\frac{5}{2}x=b \\ 2y+5x=2b \\ 5x+2y=2b \end{gathered}[/tex]
The only option with "5x + 2y" on the left side is the third option.
