How do you convert #y=(x-5)^2+3+(y-7)^2# into polar form?

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Answer 1 · 2018-08-08T06:56:48

#r^2 - r sqrt29 cos ( theta - 68.2^o ) + 3 = 0#.

The given equation represents a circle. Reorganized, it becomes

#( x - 5 )^2 + ( y - 15/2 )^2 = (sqrt17/2)^2#. See graph.
graph{( x - 5 )^2 + ( y - 15/2 )^2 - (sqrt17/2)^2=0[0 20 0 10]}

Just conversion does not reveal all this information, and as such,

does not make students understand the problems for which the

the polar frame has relative merits. In this universe, everything is an

orbiter about another orbiter, and polar frame befits rotations and

revolutions. So, I get polar equation, after knowing that the graph is

a circle.

The polar equation of the circle,

with center at #C ( c, alpha )# and radius a is

#r^2 - 2 r c cos ( theta - alpha ) + c^2 - a^2 = 0#.

Here,

Cartesian #C ( x, y ) = C ( 3, 15/2 )# and

polar #C ( c, alpha ) = C ( sqrt 29 /2, arctan ( (15/2)/3) )# and

#a = 1/2sqrt17#.

Answer:

#r^2 - r sqrt29 cos ( theta - arctan (5/2) ) + 3 = 0#.

#arctan 2.5 = 68.2^o = 1.19 rad#, nearly.