The polar equation of the circle with centre | vedclass

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(C) The Cartesian coordinates of the centre $(h, k)$ are given by $h = r_0 \cos 	heta_0$ and $k = r_0 \sin 	heta_0$,where $(r_0, 	heta_0) = (2, \fr

Vedclass · Accepted Answer

$r^2-4r \sin 	heta=5$

Vedclass · Answer

(C) The Cartesian coordinates of the centre $(h, k)$ are given by $h = r_0 \cos 	heta_0$ and $k = r_0 \sin 	heta_0$,where $(r_0, 	heta_0) = (2, \frac{\pi}{2})$.Thus,$h = 2 \cos(\frac{\pi}{2}) = 0$ and $k = 2 \sin(\frac{\pi}{2}) = 2$.The centre is $(0, 2)$ and the radius is $a = 3$.The Cartesian equation of the circle is $(x - h)^2 + (y - k)^2 = a^2$,which becomes $(x - 0)^2 + (y - 2)^2 = 3^2$.This simplifies to $x^2 + y^2 - 4y + 4 = 9$,or $x^2 + y^2 - 4y = 5$.Using the polar conversions $x^2 + y^2 = r^2$ and $y = r \sin 	heta$,we substitute into the equation:$r^2 - 4(r \sin 	heta) = 5$.Therefore,the polar equation is $r^2 - 4r \sin 	heta = 5$.

The polar equation of the circle with centre $\left(2, \frac{\pi}{2}\right)$ and radius $3$ units is :

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