Diametric Form of a Circle - Practice Questions & MCQ

Diametric Form of a Circle: 

The equation of circle, when endpoints $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ of a diameter are given, is

$
\left(\mathrm{x}-\mathrm{x}_1\right)\left(\mathrm{x}-\mathrm{x}_2\right)+\left(\mathrm{y}-\mathrm{y}_1\right)\left(\mathrm{y}-\mathrm{y}_2\right)=0
$
 

Proof:

$P(x, y)$ is any point on the circle

$
\begin{aligned}
& \text { Slope of } \mathrm{AP}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{x}-\mathrm{x}_1} \\
& \text { Slope of } \mathrm{BP}=\frac{\mathrm{y}-\mathrm{y}_2}{\mathrm{x}-\mathrm{x}_2} \\
& \because \angle \mathrm{APB}=90^{\circ} \\
& \therefore \text { Slope of AP } \times \text { Slope of } \mathrm{BP}=-1 \\
& \Rightarrow\left(\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{x}-\mathrm{x}_1}\right) \times\left(\frac{\mathrm{y}-\mathrm{y}_2}{\mathrm{x}-\mathrm{x}_2}\right)=-1 \\
& \Rightarrow\left(\mathrm{x}-\mathrm{x}_1\right)\left(\mathrm{x}-\mathrm{x}_2\right)+\left(\mathrm{y}-\mathrm{y}_1\right)\left(\mathrm{y}-\mathrm{y}_2\right)=0
\end{aligned}
$