Show that the point (x, y) given by x= 2at 1+t^2 and y=a ( 1-t^2 1+t^2 )2 lies on a circle for all real values of t such that -1 1, where a is any given real number.

x= 2at 1+t^2 ,y=a ( 1-t^2 1+t ) x^2+y^2= 4a^2t^2 (1+t^2)^2 + a^2(1-t^2)^2 (1+t^2)^2 = 4a^2t^2+a^2(1-2t^2+a^2)t^4 (1+t^2)^2 = 4a^2t^2+a^2-2a^2t^2+a^2t^4 (1+t^2)^2 = 2 a^2t^2+a^2+a^2t^2 (1+t^2)^2 = a^2(1+2t^2+t^2) (1+t^2)^2 x^2+y^2=a^2 is equation of a circle.

Show that the point (x, y) given by x= 2at 1+t^2 and y=a ( 1-t^2 …

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Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.

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Answer

$\text{x}=\frac{2\text{at}}{1+\text{t}^2},\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}}\Big)$ $\text{x}^2+\text{y}^2=\frac{4\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}+\frac{\text{a}^2(1-\text{t}^2)^2}{(1+\text{t}^2)^2}$ $=\frac{4\text{a}^2\text{t}^2+\text{a}^2(1-2\text{t}^2+\text{a}^2)\text{t}^4}{(1+\text{t}^2)^2}$ $=\frac{4\text{a}^2\text{t}^2+\text{a}^2-2\text{a}^2\text{t}^2+\text{a}^2\text{t}^4}{(1+\text{t}^2)^2}$ $=\frac{2\text{a}^2\text{t}^2+\text{a}^2+\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}$ $=\frac{\text{a}^2(1+2\text{t}^2+\text{t}^2)}{(1+\text{t}^2)^2}$ $\text{x}^2+\text{y}^2=\text{a}^2$ is equation of a circle.

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