Question

find the equation of circle whose end points of the diameter are A(-2,3) and B(3,-5)

Answer 1

$$\begin{gathered} \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circle}=\left(\frac{-2+3}{2},\frac{3-5}{2}\right)\mathrm{i}.\mathrm{e}.\left(\frac{1}{2},-1\right) \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{circle}=\sqrt{{\left(-1-3\right)}^2+{\left(\frac{1}{2}+2\right)}^2}=\frac{\sqrt{89}}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}: \\ {\left(\mathrm{x}-\frac{1}{2}\right)}^2+{\left(\mathrm{y}+1\right)}^2={\left(\frac{\sqrt{89}}{2}\right)}^2 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{4}-\mathrm{x}+{\mathrm{y}}^2+1+2\mathrm{y}=\frac{89}{4} \\ \Rightarrow \frac{4{\mathrm{x}}^2+1-4\mathrm{x}+4{\mathrm{y}}^2+4+8\mathrm{y}}{4}=\frac{89}{4} \\ \Rightarrow 4{\mathrm{x}}^2+1-4\mathrm{x}+4{\mathrm{y}}^2+4+8\mathrm{y}=89 \\ \Rightarrow 4{\mathrm{x}}^2+4{\mathrm{y}}^2-4\mathrm{x}+8\mathrm{y}-84=0 \end{gathered}$$