Circles in Polar Form Lesson

Equations of Circles in Polar Form Centered at the Origin

In this lesson, we will focus on converting circles between rectangular and polar forms. Additionally, we will learn how to graph circles on the polar grid.
Let's begin by thinking about the equation of a circle centered at the origin. $$x^2 + y^2=a^2$$ Typically, we use r to represent the radius of a circle but here we will use a since r is used with polar coordinates.
Recall with polar equations, we will replace x and y: $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ Additionally, we can replace $r^{2}$: $$r^{2}=x^{2}+ y^{2}$$ Let's convert our rectangular equation to a polar equation: $$x^{2}+ y^{2}=a^{2}$$ Replace $x^{2}+ y^{2}$: $$r^{2}=a^{2}$$ $$r=\pm a$$ Note: r can be negative in polar coordinates.
Let's look at an example.
Example #1: Find the polar equation and sketch the graph. $$x^2 + y^2=25$$ Plug into the formula given: $$x^2 + y^2=a^2$$ $$r=\pm a$$ Since a² is 25, we can write this as: $$r=\pm 5$$ Additionally, we can also do this the long way.
Replace $x^{2}+ y^{2}$: $$r^2=25$$ $$r=\pm 5$$ The negative r-value here is often a bit confusing for students. Think about it this way. Starting from the pole, no matter which angle you choose or which way you are pointing, you can walk 5 units forwards or 5 units backwards to get to a point on the circle. Therefore, the graphs of r = 5 and r = -5 coincide. Example #2: Sketch the graph of the polar equation and convert to rectangular form. $$r=\pm 6$$ Let's begin by sketching our graph: How can we convert this circle over to rectangular form?
We can reverse our formula. $$x^2 + y^2=a^2$$ $$r=\pm a$$ Since r is plus or minus 6, this means our rectangular form will be: $$x^2 + y^2=(\pm 6)^{2}$$ $$x^2 + y^2=36$$ Additionally, we can also do this the long way. $$r=\pm 6$$ Square both sides: $$r^2=36$$ Replace $r^2$: $$x^2 + y^2=36$$

Circles in Polar Form Not Centered at the Origin

In most cases, the circles we work with are not centered at the origin. How can we derive a formula for finding the polar form? To find the polar equation for a circle with radius a, let's consider the following image above:
Center: P₀ with polar coordinates given as: $$(r_{0}, θ_{0})$$ Point on the Circle: P with polar coordinates given as: $$(r, θ)$$ We form the triangle OP₀P and use the law of cosines since we have SAS. This gives us: $$a^{2}=(r_{0})^2 + (r)^{2}- 2(r_{0})(r) \cdot \text{cos}(θ - θ_{0})$$ Note: (θ - θ₀) gives us the angle opposite of side a.
In most cases, the problems given in this section involve a circle that will pass through the origin. When this occurs, we can come up with some formulas that make our life a lot easier.

Polar Equations for Circles Through the Origin Centered on the x-axis

Let's now take our formula and apply it to a circle that is centered on the x-axis and passes through the origin. $$a^{2}=(r_{0})^2 + (r)^{2}- 2(r_{0})(r) \cdot \text{cos}(θ - θ_{0})$$ If the circle passes through the origin, then r₀ will be equal to a.
Recall r₀ is the distance from the origin to the center of the circle (here this is equal to the length of the radius).
This can be used to simplify our equation:
Plug in a for r₀: $$a^{2}=a^{2}+ r^{2}- 2ar \cdot \text{cos}(θ - θ_{0})$$ Since a² is on both sides, we can remove it from the equation: $$0=r^{2}- 2(a)(r) \cdot \text{cos}(θ - θ_{0})$$ Move r² to the left side and multiply both sides by -1: $$r^{2}=2ar \cdot \text{cos}(θ - θ_{0})$$ If the circle's center lies on the positive x-axis, θ₀ is 0 and our equation becomes: $$r^{2}=2ar \cdot \text{cos}(θ - 0)$$ $$r^{2}=2ar \cdot \text{cos}\hspace{.1em}θ$$ Divide both sides by r: $$r=2a \cdot \text{cos}\hspace{.1em}θ$$ We can also come up with this formula from rectangular form. Recall the equation for a circle with a center (h, k): $$(x - h)^{2}+ (y - k)^{2}=r^{2}$$ In this case, we will swap out our r for an a since r will be used with the polar form: $$(x - h)^{2}+ (y - k)^{2}=a^{2}$$ From here, we know the center occurs at (a, 0), let's replace this in our equation: $$(x - a)^{2}+ (y - 0)^{2}=a^{2}$$ To put this equation in polar form, let's expand the binomial on the left: $$x^{2}- 2ax + a^{2}+ y^2=a^{2}$$ Get rid of the a² on each side: $$x^{2}- 2ax + y^2=0$$ Recall the definition of $r^{2}$: $$r^2=x^2 + y^2$$ Regroup: $$x^{2}+ y^2 - 2ax=0$$ Replace $x^2 + y^2$: $$r^{2}- 2ax=0$$ Let's add 2ax to each side: $$r^{2}=2ax$$ Replace x with r cos θ: $$r^{2}=2ar \cdot \text{cos}\hspace{.1em}θ$$ Divide both sides by r: $$r=2a \cdot \text{cos}\hspace{.1em}θ$$ Let's look at example.
Example #3: Convert to polar form and sketch the graph of the polar equation. $$(x + 2)^{2}+ y^{2}=4$$ We should notice that this example matches the pattern given above. $$(x - a)^{2}+ y^{2}=a^{2}$$ Here, a is (-2): $$(x - (-2))^{2}+ y^{2}=(-2)^{2}$$ $$(x + 2)^{2}+ y^{2}=4$$ The circle has a center (-2, 0).
To convert to polar, we can just plug into the formula: $$r=2a \cdot \text{cos}\hspace{.1em}θ$$ Here, a = -2: $$r=-4 \cdot \text{cos}\hspace{.1em}θ$$ We can also find this the long way. $$(x + 2)^{2}+ y^{2}=4$$ Expand first: $$x^{2}+ 4x + 4 + y^{2}=4$$ Simplify: $$x^{2}+ 4x + y^{2}=0$$ Rearrange Terms: $$x^{2}+ y^{2}=-4x$$ Replace $x^{2}+ y^{2}$: $$r^{2}=-4x$$ Replace x with r cos θ: $$r^{2}=-4r \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Divide both sides by r: $$r=-4\cdot \text{cos}\hspace{.1em}θ$$ Example #4: Sketch the graph of the polar equation and convert to rectangular form. $$r=6 \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Let's consider our two forms: $$(x - a)^{2}+ y^{2}=a^{2}$$ $$r=2a \cdot \text{cos}\hspace{.1em}θ$$ Here, 2a is 6, so a must be 3: $$r=2(3) \hspace{.1em}\text{cos}\hspace{.1em}θ$$ To convert this over: $$(x - 3)^{2}+ y^{2}=3^{2}$$ $$(x - 3)^{2}+ y^{2}=9$$ Again, we can do this the long way. $$r=6 \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Multiply both sides by r: $$r^{2}=6r \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Replace $r^{2}$: $$x^{2}+ y^{2}=6r \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Replace r cos θ: $$x^{2}+ y^{2}=6x$$ Subtract 6x from both sides: $$x^{2}- 6x + y^{2}=0$$ Complete the square: $$x^{2}- 6x + 9 + y^{2}=9$$ Factor the perfect square trinomial: $$(x - 3)^{2}+ y^{2}=9$$

Polar Equations for Circles Through the Origin Centered on the y-axis

Similarly, we will also see circles that are centered on the y-axis and pass through the origin. $$r^{2}=2ar \cdot \text{cos}(θ - θ_{0})$$ If the circle's center lies on the positive y-axis, θ₀ is 90° and our equation becomes: $$r^{2}=2ar \cdot \text{cos}(θ - 90°)$$ From here, let's use our difference identity for cosine: $$\text{cos}(θ - 90°)$$ $$=\text{cos}\hspace{.1em}θ \cdot \text{cos}\hspace{.1em}90° + \text{sin}\hspace{.1em}θ \cdot \text{sin}90°$$ $$=\text{sin}\hspace{.1em}θ$$ $$r^{2}=2ar \cdot \text{sin}\hspace{.1em}θ$$ Divide both sides by r: $$r=2a \cdot \text{sin}\hspace{.1em}θ$$ Again, we can also come across this from rectangular form: $$(x - h)^{2}+ (y - k)^{2}=r^{2}$$ If we replace our center with (0, a) and our radius r with a, we obtain: $$x^{2}+ (y - a)^{2}=a^2$$ Expand: $$x^{2}+ y^{2}- 2ay + a^{2}=a^{2}$$ Simplify and rearrange terms: $$x^{2}+ y^{2}=2ay$$ Replace $x^2 + y^2$ and y: $$r^{2}=2ar \cdot \text{sin}\hspace{.1em}θ$$ Divide both sides by r: $$r=2a \cdot \text{sin}\hspace{.1em}θ$$ Example #5: Sketch the graph of the polar equation and convert to rectangular form. $$r=-4 \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Let's consider our two forms: $$r=2a \cdot \text{sin}\hspace{.1em}θ$$ $$x^{2}+ (y - a)^{2}=a^2$$ $$r=-4 \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Here, 2a is -4, so a must be -2: $$r=2(-2) \hspace{.1em}\text{cos}\hspace{.1em}θ$$ To convert this over: $$x^{2}+ (y - (-2))^{2}=(-2)^2$$ $$x^{2}+ (y + 2)^{2}=4$$ Again, we can always do this the long way: $$r=-4 \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Multiply both sides by r: $$r^{2}=-4r \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Replace $r^{2}$: $$x^{2}+ y^{2}=-4r \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Replace the r sin θ with y: $$x^{2}+ y^{2}=-4y$$ Add 4y to both sides and complete the square: $$x^{2}+ y^{2}+ 4y + 4=4$$ Factor the perfect square trinomial: $$x^{2}+ (y + 2)^{2}=4$$ Example #6: Convert to polar form and sketch the graph of the polar equation. $$x^{2}+ (y - 4)^{2}=16$$ We should notice that this example matches the pattern given above. $$r=2a \cdot \text{sin}\hspace{.1em}θ$$ $$x^{2}+ (y - a)^{2}=a^2$$ $$x^{2}+ (y - 4)^{2}=4^2$$ The circle has a center at (0, 4).
To convert to polar, we can just plug into the formula: $$r=2a \cdot \text{sin}\hspace{.1em}θ$$ Here a = 4: $$r=8 \cdot \text{sin}\hspace{.1em}θ$$ Again, we can always do this the long way: $$x^{2}+ (y - 4)^{2}=16$$ Expand: $$x^{2}+ y^{2}- 8y + 16=16$$ Simplify: $$x^{2}+ y^{2}- 8y=0$$ Replace $x^{2}+ y^{2}$ with $r^{2}$: $$r^{2}- 8y=0$$ Add 8y to both sides: $$r^{2}=8y$$ Replace y with r sin θ: $$r^{2}=8r \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Divide both sides by r: $$r=8 \hspace{.1em}\text{sin}\hspace{.1em}θ$$

Circle Centered at (a, b) with Radius $\sqrt{a^{2}+ b^{2}}$

Lastly, let's discuss one other special type of circle that goes through the origin. In this case, our center will not be on the x or y axis, however, our circle will still pass through the origin. Using a right triangle, we can show that the distance from the origin to the center of the circle is: $$\sqrt{a^{2}+ b^{2}}$$ We can conclude that when a circle passes through the origin and has a center (a, b), our radius will be given as: $$\sqrt{a^{2}+ b^{2}}$$ Let's plug into our standard form of a circle: $$(x - a)^{2}+ (y - b)^{2}=\left(\sqrt{a^{2}+ b^{2}}\right)^{2}$$ Simplify: $$(x - a)^{2}+ (y - b)^{2}=a^{2}+ b^{2}$$ Expand the left side: $$x^{2}- 2ax + a^{2}+ y^{2}- 2by + b^{2}=a^{2}+ b^{2}$$ Simplify: $$x^{2}- 2ax + y^{2}- 2by=0$$ Regroup: $$x^{2}+ y^{2}=2ax + 2by$$ Replace $x^{2}+ y^{2}$: $$r^{2}=2ax + 2by$$ Replace x with r cos θ and y with r sin θ: $$r^{2}=2ar \cdot \text{cos}\hspace{.1em}θ + 2b r \cdot \text{sin}\hspace{.1em}θ$$ Factor out the r on the right side: $$r^{2}=r(2a \cdot \text{cos}\hspace{.1em}θ + 2b \cdot \text{sin}\hspace{.1em}θ)$$ Divide both sides by r: $$r=2a \cdot \text{cos}\hspace{.1em}θ + 2b \cdot \text{sin}\hspace{.1em}θ$$ Let's look at a few examples.
Example #7: Sketch the graph of the polar equation and convert to rectangular form. $$r=6 \hspace{.1em}\text{cos}\left(θ + \frac{π}{3}\right)$$ To clean up the right side, let's use the cosine sum identity: $$6\left[\text{cos}\left(θ + \frac{π}{3}\right)\right]$$ $$=6\left[\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}\hspace{.1em}\frac{π}{3}- \text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}\frac{π}{3}\right]$$ $$=6\left[\frac{1}{2}\cdot \text{cos}\hspace{.1em}θ - \frac{\sqrt{3}}{2}\cdot \text{sin}\hspace{.1em}θ\right]$$ $$=3 \hspace{.1em}\text{cos}\hspace{.1em}θ - 3\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}θ$$ Let's return: $$ r=3 \hspace{.1em}\text{cos}\hspace{.1em}θ - 3\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}θ$$ Notice that this matches our pattern from above: $$r=2a \cdot \text{cos}\hspace{.1em}θ + 2b \cdot \text{sin}\hspace{.1em}θ$$ $$(x - a)^{2}+ (y - b)^{2}=a^{2}+ b^{2}$$ Here, 2a is 3: $$2a=3$$ $$a=\frac{3}{2}$$ Here, 2b is $-3\sqrt{3}$: $$2b=-3\sqrt{3}$$ $$b=-\frac{3\sqrt{3}}{2}$$ We can just plug into the formula: $$(x - a)^{2}+ (y - b)^{2}=a^{2}+ b^{2}$$ $$\left(x - \frac{3}{2}\right)^{2}+ \left(y + \frac{3\sqrt{3}}{2}\right)^{2}=\frac{9}{4}+ \frac{27}{4}$$ $$\left(x - \frac{3}{2}\right)^{2}+ \left(y + \frac{3\sqrt{3}}{2}\right)^{2}=9$$ Again, we can also do this the long way: $$ r=3 \hspace{.1em}\text{cos}\hspace{.1em}θ - 3\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}θ$$ Multiply both sides by r: $$r^{2}=3r \hspace{.1em}\text{cos}\hspace{.1em}θ - 3r\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}θ$$ Replace r cos θ with x, r sin θ with y: $$r^{2}=3x - 3y\sqrt{3}$$ Replace $r^{2}$ with $x^{2}+ y^{2}$: $$x^{2}+ y^{2}=3x - 3y\sqrt{3}$$ We need to complete the square: $$(x^{2}- 3x) + (y^{2}+ 3y\sqrt{3})=0$$ $$\left(x^{2}- 3x + \frac{9}{4}\right) + \left(y^{2}+ 3y\sqrt{3}+ \frac{27}{4}\right)=\frac{9}{4}+ \frac{27}{4}$$ $$\left(x - \frac{3}{2}\right)^{2}+ \left(y + \frac{3 \sqrt{3}}{2}\right)^2=\frac{36}{4}$$ $$\left(x - \frac{3}{2}\right)^{2}+ \left(y + \frac{3 \sqrt{3}}{2}\right)^2=9$$ Example #8: Convert to polar form and sketch the graph of the polar equation. $$(x + 3)^{2}+ (y + 1)^{2}=10$$ Does this match our pattern? Check: $$(x - a)^{2}+ (y - b)^{2}=a^{2}+ b^{2}$$ $$(x - (-3))^{2}+ (y - (-1))^{2}=(-3)^{2}+ (-1)^{2}$$ $$(x + 3)^{2}+ (y + 1)^{2}=9 + 1$$ $$(x + 3)^{2}+ (y + 1)^{2}=10$$ Since this matches our pattern, we can use our shortcut. $$r=2a \cdot \text{cos}\hspace{.1em}θ + 2b \cdot \text{sin}\hspace{.1em}θ$$ Here, a is -3, b is -1: $$r=2(-3) \cdot \text{cos}\hspace{.1em}θ + 2(-1) \cdot \text{sin}\hspace{.1em}θ$$ $$r=-6 \cdot \text{cos}\hspace{.1em}θ - 2 \cdot \text{sin}\hspace{.1em}θ$$ Again, we can always do this the long way. $$(x + 3)^{2}+ (y + 1)^{2}=10$$ Expand: $$x^2 + 6x + 9 + y^2 + 2y + 1=10$$ Simplify and rearrange terms: $$x^2 + y^2=-6x - 2y$$ Replace $x^2 + y^2$: $$r^2=-6x - 2y$$ Replace x with r cos θ and y with r sin θ: $$r^2=-6r \cdot \text{cos}\hspace{.1em}θ - 2r \cdot \text{sin}\hspace{.1em}θ$$ Factor out the r on the right side: $$r^2=r(-6 \cdot \text{cos}\hspace{.1em}θ - 2 \cdot \text{sin}\hspace{.1em}θ)$$ Divide each side by r: $$r=-6 \cdot \text{cos}\hspace{.1em}θ - 2 \cdot \text{sin}\hspace{.1em}θ$$

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