# Angle in Semicircle is Right Angle

Consider the equation of the circle with centre at the origin $O\left( {0,0} \right)$ is given by the equation

Let $AOB$ be nay diameters of the circle and $P\left( {{x_2},{y_2}} \right)$ be any point on the given circle.
We shall show that $m\angle APB = {90^ \circ }$.
Suppose that coordinates of $A$ is $\left( {{x_1},{y_1}} \right)$, then $B$ has coordinates $\left( { - {x_1}, - {y_1}} \right)$.
Slope of $AP = \frac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}} = {m_1}$
Slope of $BP = \frac{{{y_1} + {y_2}}}{{{x_1} + {x_2}}} = {m_2}$
Now multiplying the slopes ${m_1}$and ${m_2}$, we get
${m_1}{m_2} = \frac{{{y_1}^2 - {y_2}^2}}{{{x_1}^2 - {x_2}^2}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)$
Since points $A\left( {{x_1},{y_1}} \right)$ and $P\left( {{x_2},{y_2}} \right)$ lies on the circle, we have

Substituting the values of ${x_1}^2$ and ${x_2}^2$ from equation (ii) into equation (i), we get

This is the condition of perpendicular lines.
Thus $AP \bot BP$ and so $m\angle APB = {90^ \circ }$.