Question

How do you convert `x^2 + y^2 = 49` into polar form?

Answer 1

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Polar form of the equation in rectangular form : `color(red)(x^2+y^2=49`

is

**`color(blue)(r=+7 or r = -7`

Explanation:

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**Given the equation in Rectangular Form: ** `color(red)(x^2+y^2=49`

Note that the Equation of the Circle : `color(blue)((x-h)^2+(y-k)^2=r^2`,

where

`color(blue)((h,k)` is the **Center ** of the Circle

and `color(blue)(r` is the **Radius ** of the Circle.

In other words:

`color(blue)(h` represents the x-offset from the origin and

`color(blue)(k` represents the y-offset from the origin.

The given equation `color(red)(x^2+y^2=49`

[ rewrite the equation as : `color(blue)(x^2+y^2=7^2` ]

represents the equation of the circle

with `color(green)("Center at "(0,0)` and `color(brown)("Radius "7`

The Polar Form of the equation can be obtained by using the formula:

`color(blue)(x^2+y^2 = r^2`

`color(blue)(x=r cos(theta)` and

`color(blue)(y=r sin(theta)`

Now, consider the given equation: `color(red)(x^2+y^2=49`

Using `color(blue)(x=r cos(theta)` and `color(blue)(y=r sin(theta)`

`rArr [r cos(theta)]^2 + [r sin(theta)]^2=49`

`rArr r^2 cos^2(theta)+r^2 sin^2(theta) = 49`

`rArr r^2[cos^2(theta)+sin^2(theta)] = 49`

Using the identity: `color(red)(sin^2(theta)+cos^2(theta)-=1`

the equation becomes:

`rArr r^2(1)=49`

`rArr r^2=49`

`rArr r=+-sqrt(49)`

`color(blue)(r = +7 or r = -7`.

This is the required Polar Form.

Graph this:
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Compare this graph with the graph below which is drawn to represent the equation of the circle:

This represents a perfect circle with radius 7 on the origin.