How do you convert #(x-1)^2 +y^2 =5# in polar form?

2 Answers
Feb 20, 2016

r = cos #theta# + #sqrt# ( 4 + cos #theta#. cos #theta#)

Explanation:

x = r cos #theta# and y = r sin #theta# transforms the equation to a quadratic in r. r is set as non-negative, with [0, 2#pi#] for .#theta#.
This is the polar equation of the circle with center in polar (1. 0 )and radius #sqrt#5. .

#r=cos theta+-sqrt(cos^2 theta+4)#

Explanation:

From the given, #(x-1)^2+y^2=5#
Use #x=r cos theta# and #y=r sin theta#

Substitute these in the given variables x and y

#(x-1)^2+y^2=5#

#(r cos theta-1)^2+(r sin theta)^2=5#

#r^2 cos^2 theta-2 r cos theta+1+r^2 sin^2 theta=5#

factor out #r^2#

#r^2(cos^2 theta+sin^2 theta)-2 r cos theta+1-5=0#
simplify
#r^2(1)-2 r cos theta-4=0#
#r^2-2 r cos theta-4=0#

Use now Quadratic Equation to solve for #r# in terms of the other variable

#r=(-b+-sqrt(b^2-4*a*c))/(2*a)#

using #a=1#, and #b=-2 cos theta#,and #c=-4#

#r=(-(-2 cos theta)+-sqrt((-2 cos theta)^2-4*1*(-4)))/(2*1)#

#r=(2cos theta+-sqrt((4 cos^2 theta+16)))/2#

which simplifies to

#r=cos theta+-sqrt(cos^2 theta+4)#

have a nice day!