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

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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.