How do you convert r = 4 cosθ + 4 sinθ into a cartesian equation?

Question

↳Redirected from "What is #log_e# of e?"

8893 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-18T17:01:00

#x^2+y^2-4(x+y)=0#

We know that #x=rcostheta# and #y=rsintheta#

#rArr x/r =costheta# and #y/r=sintheta#

Put the above values in the equation :-

#r=4costheta +4sintheta#

#rArrr=(4(x+y))/r#

#rArrr^2=4(x+y)#

Also we know that #x^2+y^2=r^2# ; put in the equation we get :-

#rArrx^2+y^2=4(x+y)#

#:.x^2+y^2-4(x+y)=0# which is the cartesian form of the given equation.