# Angle in Semicircle is a Right Angle

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

\[{x^2} + {y^2} = {r^2}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]

Let $$AOB$$ be any diameter 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 $$A$$ are $$\left( {{x_1},{y_1}} \right)$$, 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 $$P\left( {{x_2},{y_2}} \right)$$ lie on the circle, we have

\[\left. \begin{gathered} {x_1}^2 + {y_1}^2 = {r^2} \Rightarrow {x_1}^2 = {r^2} – {y_1}^2 \\ {x_2}^2 + {y_2}^2 = {r^2} \Rightarrow {x_2}^2 = {r^2} – {y_2}^2 \\ \end{gathered} \right\}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)\]

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

\[{m_1}{m_2} = \frac{{{y_1}^2 – {y_2}^2}}{{\left( {{r^2} – {y_1}^2} \right) – \left( {{r^2} – {y_2}^2} \right)}} = \frac{{{y_1}^2 – {y_2}^2}}{{ – \left( {{y_1}^2 – {y_2}^2} \right)}} = – 1\]

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