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


angle-semi-circle

{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 }.