Question

How do you convert $ r = 2\sin \theta $ into cartesian form?

Answer 1

Hint: In the cartesian coordinate system, a point is represented as $ (x,y) $ where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis. Points of the form $ (r,\theta ) $ are called polar coordinates, where r is the distance of the point from the origin and $ \theta $ is the counter-clockwise angle between the line joining the point and the origin, and the x-axis. A right-angled triangle is formed by x, y and r, where r is the hypotenuse, x is the base and y is the height of the triangle, so by Pythagoras theorem, we have - $ {x^2} + {y^2} = {r^2} $ and by trigonometry we have - $ \sin \theta = \dfrac{{perpendicular}}{{hypotenuse}} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} $ . In this question, we have to convert a polar equation into the cartesian form. Using the above information, we will express all the quantities in terms of x and y and by further solving the equation, we get the cartesian form.

Complete step-by-step answer:
Given,
$ r = 2\sin \theta $
We know that –
 $
  {r^2} = {x^2} + {y^2} \\
   \Rightarrow r = \sqrt {{x^2} + {y^2}} \;
  $
And $ \sin \theta = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} $
Using the above two values in the given polar equation, we get –
 $
  \sqrt {{x^2} + {y^2}} = 2(\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}) \\
   \Rightarrow {x^2} + {y^2} = 2y \\
   \Rightarrow {x^2} + {y^2} - 2y = 0 \;
  $
Adding 1 on both sides of the above equation, we get –
 $
  {x^2} + {y^2} - 2y + 1 = 1 \\
  {x^2} + {y^2} - 2 \times 1 \times y + {(1)^2} = 1 \\
   \Rightarrow {x^2} + {(y - 1)^2} = 1 \;
  $
Hence the given polar equation is written in the cartesian form as $ {x^2} + {(y - 1)^2} = 1 $ .
So, the correct answer is “ $ {x^2} + {(y - 1)^2} = 1 $ ”.

Note: The given polar equation represents a circle as r is the distance of the point from the origin, it is the radius of the circle and by putting different values of $ \theta $ , we get different points lying on the circle. So, the obtained cartesian equation is the equation of the circle. On comparing this equation with the general equation of the circle $ (x - h) + {(y - k)^2} = {r^2} $ , we see that the coordinates of the centre of this circle is $ (0,1) $ and the radius of the circle is 1 unit.