Determine Polar Equation...1

[nycmathguy]

Can you work out 36?

 

[MathLover1]

polar equation of the circle

radius R=sqrt(6) and
polar coordinates of the center:
a. (2,pi)
b. (2,3pi/4)
c. (0,0)

The equation of a circle with (h, k) center and r radius is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation.

a.
given center at (2,π)=>h=2, k=π and radius (let it be) R=sqrt(6)

(x-2)^2 + (y-π )^2 = (sqrt(6))^2
(x-2)^2 + (y-π )^2 = 6

To find the polar form of equation of a circle, replace the value of x = r cos θ and y = r sin θ,
Hence, we get:

(r*cos (θ)-2)^2 + (r*sin (θ)-π)^2 = 6

do this your self
b. (2,3pi/4)
c. (0,0)

 

[nycmathguy]

polar equation of the circle

radius R=sqrt(6) and
polar coordinates of the center:
a. (2,pi)
b. (2,3pi/4)
c. (0,0)

The equation of a circle with (h, k) center and r radius is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation.

a.
given center at (2,π)=>h=2, k=π and radius (let it be) R=sqrt(6)

(x-2)^2 + (y-π )^2 = (sqrt(6))^2
(x-2)^2 + (y-π )^2 = 6

To find the polar form of equation of a circle, replace the value of x = r cos θ and y = r sin θ,
Hence, we get:

(r*cos (θ)-2)^2 + (r*sin (θ)-π)^2 = 6

do this your self
b. (2,3pi/4)
c. (0,0)

Easy to understand this reply. I will do the rest when time allows.

 

[nycmathguy]

polar equation of the circle

radius R=sqrt(6) and
polar coordinates of the center:
a. (2,pi)
b. (2,3pi/4)
c. (0,0)

The equation of a circle with (h, k) center and r radius is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation.

a.
given center at (2,π)=>h=2, k=π and radius (let it be) R=sqrt(6)

(x-2)^2 + (y-π )^2 = (sqrt(6))^2
(x-2)^2 + (y-π )^2 = 6

To find the polar form of equation of a circle, replace the value of x = r cos θ and y = r sin θ,
Hence, we get:

(r*cos (θ)-2)^2 + (r*sin (θ)-π)^2 = 6

do this your self
b. (2,3pi/4)
c. (0,0)

 

[MathLover1]

correct

 

[nycmathguy]

correct

This is a nice, basic exercise.