The centre and radius of a circle x=4 aleft(f | vedclass

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(B) The given parametric equations are $x = 4a \left(\frac{1-t^2}{1+t^2}ight)$ and $y = \frac{8at}{1+t^2}$.Let $t = 	an 	heta$. Then $\cos 2	heta

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$(0,0)$ and $4 a$ units

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(B) The given parametric equations are $x = 4a \left(\frac{1-t^2}{1+t^2}ight)$ and $y = \frac{8at}{1+t^2}$.Let $t = 	an 	heta$. Then $\cos 2	heta = \frac{1-t^2}{1+t^2}$ and $\sin 2	heta = \frac{2t}{1+t^2}$.Substituting these,we get $x = 4a \cos 2	heta$ and $y = 4a \sin 2	heta$.Squaring and adding both equations:$x^2 + y^2 = (4a \cos 2	heta)^2 + (4a \sin 2	heta)^2$$x^2 + y^2 = 16a^2 (\cos^2 2	heta + \sin^2 2	heta)$$x^2 + y^2 = (4a)^2$.This represents a circle with centre $(0,0)$ and radius $4a$.

The centre and radius of a circle $x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}$ are respectively:

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