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 \[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\].