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


angle-semi-circle

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


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

\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 }.

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