Get help with circle in polar coordinates concept in this page and understand the concept better with workout examples provided on polar coordinates of circle. For more help students can get connected to an online tutor and get the required help with polar coordinates circle. First let's get introduced with the concept of polar coordinates system.

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Polar coordinates is another form of coordinates instead of Cartesian coordinates. In Cartesian coordinates, any point is determined by (x,y) a unique position for the point. Here, in Polar coordinates, any point is uniquely determined by two coordinates, r and $\theta$. The r is the distance of the point from the origin and $\theta$ is the angle the line joining the point and the origin makes with the horizontal line.

Hence in polar coordinates, which is other way of expressing points in ordered pairs, r, $\theta$ are used instead of x, y where r represents the distance of the point from origin and $\theta$ the angle the line joining point and origin makes with horizontal axis.

**Circle:** A circle is the locus of points which maintain equal distance from a fixed point. The fixed point is called the centre of the circle and the distance from centre to any point in the circle is called radius.

**Circle polar coordinates:** Now we see how we can write equation of circle in polar coordinates and how to find out other aspects about circle in polar coordinates.

The radial coordinate is denoted by *r*, and the angular coordinate by $\theta$

Since the circle has equal distance from centre, if origin is the centre, circle will have points in the circle represented by

(rcos$\theta$, rsin$\theta$). This ordered pair for r fixed and $\theta$ varying from 0 to 360 degrees form a circle because the distance is always r with the origin. So in polar coordinates equation of the circle is of the form (rcos$\theta$, rsin$\theta$) where r is fixed but theta varying from 0 to 360 degrees.

**When (g,f) is the centre instead of origin:**

The equation of the circle when we shift origin to (g, f) is (rcos $\theta$, rsin $\theta$). Now by shifting back the origin to (0, 0) we get equation for points in a circle with centre (g, f) as **(g + rcos, f + rsin$\theta$)**

Polar equation of a circle with equation $x^2 + y^2 = r^2$ can be given using $(rcos \theta, rsin\theta)$ where $\theta = tan^{-1}\frac{y}{x}$. To find if a point falls on a circle using its polar coordinates, we need to check whether the value of the angle $\theta$ will come between 0 and 2$\pi$.

**1) **Find the equation of the circle with centre at (2, -3) and radius 4.

**Sol:** We know the ordered pair for the circle is of the form **(g + rcos, f + rsin ****$\theta$).**

** So **here substituting for g,f and r we get, (2 + 4cos $\theta$, -3 + 4sin $\theta$) for 0 $\leq \theta \leq$ 2 $\pi$.

**Relation between polar and cartesian coordinates:**

The relation is given by x=rcos $\theta$,y=rsin $\theta$.

and r = $\sqrt {x ^2 + y^2}$ and $\theta$ = tan $^{-}(\frac{y}{x})$.

**2) **Find the parametric equation of the circle 4x$^2$ + 4y$^2$ =9.

**Sol:** Dividing by 4, we get x$^2$ + y$^2$ = $\frac{9}{4}$.

So r = $\frac{3}{2}$. In polar coordinates the circle is represented by ($\frac{3}{2}$cos $\theta$, $\frac{3}{2}$sin$\theta$) for 0 $\leq$ $\theta$ $\leq$ 2 $\pi$.

**3) Find **Cartesian equation of the circle whose parametric equations are x = $\frac{1}{4}$cos$\theta$,$\frac{1}{4}$sin$\theta$ for 0 $\leq$ $\theta$ $\leq$2$\pi$.

**Sol:** We know this equation represents a circle where origin is centre and radius is $\frac{1}{4}$.

So equation in Cartesian coordinates is x$^2$ + y$^2$ = $\frac{1}{16}$.

Here are few more solved problems on polar coordinates circle provided for a better understanding:

**Problem 1:** If (4,1) is one point on a circle represented by (1 + 5cos$\theta$,- 3 + 5sin $\theta$) find the other extremity of the diameter for this point.

**Sol:** We know for the other extremity of the diameter is increased by 180 degrees.

Now for 4=1 + 5cos $\theta$ We have cos$\theta$ = $\frac{3}{5}$. Similarly 1= -3 + 5 sin $\theta$ We get sin $\theta$ =$\frac{4}{5}$.

So for other extremity we have cos ($\theta$ +180) = -cos$\theta$= - $\frac{3}{5}$ and sin($\theta$ + 180) = -sin $\theta$ = - $\frac{4}{5}$.

So the other point is (1 + 5 $\times$ - $\frac{3}{5}$,-3 + 5 $\times$ $\frac{4}{5}$) = (-2, -7)

**Problem 2**: If a circle has radius 3, and centre (2,-3) prove that it passes through (2,0)

**Sol:** The equation of the circle is (2 + 3cos $\theta$, -3 + 3sin $\theta$)

To prove that the circle passes through (2,0) we should prove that 2 =2 + 3$cos \theta$ and 0 = - 3 + 3$sin \theta$ for some theta between 0 and 2 $\pi$.

To prove that cos$\theta$ =0 and sin$\theta$ =1 for some $\theta$ which is true when $\theta$ =90 degrees,

Hence the circle passes through that point.

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