Question

The polar equation of the circle with center $\left( {2,\dfrac{\pi }{2}} \right)$ and radius 3 units is:
A. ${r^2} + 4r\cos \theta = 5$
B. ${r^2} + 4r\sin \theta = 5$
C. ${r^2} - 4r\sin \theta = 5$
D. ${r^2} - 4r\cos \theta = 5$

Answer 1

Hint : A polar equation is any equation that describes a relation between $r$ and $\theta$, where $r$ represents the distance from the pole (origin) to a point on a curve, and $\theta$ represents the counterclockwise angle made by a point on a curve, the pole, and the positive x-axis.

Complete step by step solution :
Given centre of the circle is $\left( {2,\dfrac{\pi }{2}} \right)$ and its radius is 3 units.
We know that the general equation of the circle in polar form is ${r^2} - 2r{r_0}\cos \left( {\theta - \gamma } \right) + {r_0}^2 = {a^2}$ where $\left( {{r_0},\gamma } \right)$ is the centre and $a$ is the radius.
So, the equation of the circle with centre $\left( {2,\dfrac{\pi }{2}} \right)$ and radius of 3 units is given by

\Rightarrow {r^2} - 2r\left( 2 \right)\cos \left( {\theta - \dfrac{\pi }{2}} \right) + {2^2} = {3^2} \\\ \Rightarrow {r^2} - 4r\cos \left\\{ { - \left( {\dfrac{\pi }{2} - \theta } \right)} \right\\} + 4 = 9 \\\ \Rightarrow {r^2} - 4r\cos \left( {\dfrac{\pi }{2} - \theta } \right) = 9 - 4 \\\ \therefore {r^2} - 4r\sin \theta = 5 \\\

Therefore, the required equation of the circle is ${r^2} - 4r\sin \theta = 5$.
Thus, the correct option is C. ${r^2} - 4r\sin \theta = 5$

Note : The general equation of the circle in polar form is ${r^2} - 2r{r_0}\cos \left( {\theta - \gamma } \right) + {r_0}^2 = {a^2}$ where $\left( {{r_0},\gamma } \right)$ is the centre and $a$ is the radius. The general equation of a circle with radius $r$ units and with centre $\left( {a,b} \right)$ in a two dimensional cartesian coordinate plane is given by ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$.