show that the point(x,y)given by x=2at/1+t² and y=a(1-t²)/1+t² lies on a circle for all real values of t such that -1<t<1 ,where a is any given real nos.
or
find the equations of the altitudes of the triangle whose vertices are A(7,-1),B(-2,8) and C(1,2)

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{x}=\frac{2\mathrm{at}}{1+{\mathrm{t}}^2}..\left(\mathrm{i}\right) \\ \mathrm{y}=\frac{\mathrm{a}\left(1-{\mathrm{t}}^2\right)}{1+{\mathrm{t}}^2}...\left(\mathrm{ii}\right) \\ \mathrm{Square} \mathrm{and} \mathrm{add} \\ {\mathrm{x}}^2+{\mathrm{y}}^2=\frac{4{\mathrm{a}}^2{\mathrm{t}}^2}{{\left(1+{\mathrm{t}}^2\right)}^2}+\frac{{\mathrm{a}}^2{\left(1-{\mathrm{t}}^2\right)}^2}{{\left(1+{\mathrm{t}}^2\right)}^2} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\left\{\frac{4{\mathrm{t}}^2+{\left(1-{\mathrm{t}}^2\right)}^2}{{\left(1+{\mathrm{t}}^2\right)}^2}\right\} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\left\{\frac{4{\mathrm{t}}^2+1+{\mathrm{t}}^4-2{\mathrm{t}}^2}{{\left(1+{\mathrm{t}}^2\right)}^2}\right\} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\left\{\frac{1+{\mathrm{t}}^4+2{\mathrm{t}}^2}{{\left(1+{\mathrm{t}}^2\right)}^2}\right\} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\left\{\frac{{\left(1+{\mathrm{t}}^2\right)}^2}{{\left(1+{\mathrm{t}}^2\right)}^2}\right\} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \mathrm{which} \mathrm{is} \mathrm{a} \mathrm{circle} \left(\mathrm{Hence} \mathrm{Proved}\right) \end{gathered}$$

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