Show that the point x-y given by x-2at1-t2 and y-a1-t21-t2 lies on a circle for all real values of t such that -1- t- 1- where a is any given real number-

We have #160; x=2at1+t^2..........(i) y=a(1 t^21+t^2).......(ii) Squaring and adding (i) and (ii) x^2+y^2=4a^2t^2(1+t^2)^2+a^2(1 t^2)^2(1+t^2)^2 ...

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Standard XII

Mathematics

Properties of Set Operation

Show that the...

Question

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

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Solution

We have
$x = \frac{2 a t}{1 + t^{2}} . . . . . . . . . . (i)$

$y = a (\frac{1 - t^{2}}{1 + t^{2}}) . . . . . . . (i i)$

Squaring and adding (i) and (ii)

$x^{2} + y^{2} = \frac{4 a^{2} t^{2}}{(1 + t^{2})^{2}} + \frac{a^{2} (1 - t^{2})^{2}}{(1 + t^{2})^{2}}$

$x^{2} + y^{2} = \frac{4 a^{2} t^{2} + a^{2} (1 - t^{2})^{2}}{(1 + t^{2})^{2}}$

$x^{2} + y^{2} = a^{2} {\frac{4 t^{2} + 1 - 2 t^{2} + t^{4}}{(1 + t^{2})^{2}}}$

$x^{2} + y^{2} = a^{2} {\frac{t^{4} + 1 + 2 t^{2}}{(1 + t^{2})^{2}}}$

$x^{2} + y^{2} = a^{2} {\frac{(1 + t^{2})^{2}}{(1 + t^{2})^{2}}}$

$x^{2} + y^{2} = a^{2}$ which is a circle
Hence proved

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Show that the point (x,y) given by x=2at1+t2 and y=a(1−t21+t2) lies on a circle for all real values of t such that −1≤t≤1, where a is any given real number.

We have x=2at1+t2..........(i)y=a(1−t21+t2).......(ii)Squaring and adding (i) and (ii)x2+y2=4a2t2(1+t2)2+a2(1−t2)2(1+t2)2x2+y2=4a2t2+a2(1−t2)2(1+t2)2x2+y2=a2{4t2+1−2t2+t4(1+t2)2}x2+y2=a2{t4+1+2t2(1+t2)2}x2+y2=a2{(1+t2)2(1+t2)2}x2+y2=a2 which is a circleHence proved

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