Solveeit Logo
Login

Question

Question: How do you convert \( r = 2\sin \theta \) into cartesian form?...

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

Explanation

Solution

Hint : In the cartesian coordinate system, a point is represented as (x,y)(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,θ)(r,\theta ) are called polar coordinates, where r is the distance of the point from the origin and θ\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 - x2+y2=r2{x^2} + {y^2} = {r^2} and by trigonometry we have - sin⁡θ=perpendicularhypotenuse=yx2+y2\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=2sin⁡θr = 2\sin \theta
We know that –
r2=x2+y2 ⇒r=x2+y2   {r^2} = {x^2} + {y^2} \\\ \Rightarrow r = \sqrt {{x^2} + {y^2}} \;
And sin⁡θ=yx2+y2\sin \theta = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}
Using the above two values in the given polar equation, we get –
x2+y2=2(yx2+y2) ⇒x2+y2=2y ⇒x2+y2−2y=0   \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 –
x2+y2−2y+1=1 x2+y2−2×1×y+(1)2=1 ⇒x2+(y−1)2=1   {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 x2+(y−1)2=1{x^2} + {(y - 1)^2} = 1 .
So, the correct answer is “ x2+(y−1)2=1{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 θ\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=r2(x - h) + {(y - k)^2} = {r^2} , we see that the coordinates of the centre of this circle is (0,1)(0,1) and the radius of the circle is 1 unit.