r = 2 cos θ + 6 sin θ

r^2 = 2r cos θ + 6r sin θ

x^2+y^2 = 2x + 6y

x^2-2x+y2-6y = 0

x^2-2x+1-1 + y^2-6y+9-9 = 0

(x+1)^2 +(y-3)^2 = 10

Circle with center (-1,3) and radius sqrt(10)

r = 6 cos θ

r^2 = r cos θ

x^2+y^2 = 6x

x^2-6x+y^2=0

(x^2-6x+9-9) + y^2 = 0

(x-3)^2+y^2 = 9

circle with center (3,0) and radius 3

Hello,

   𝑟 = 2·cos(𝜃) + 6·sin(𝜃)

   𝑟² = 2𝑟·cos(𝜃) + 6𝑟·sin(𝜃)       ←←← Multiply everything by 𝑟

   𝑟² = 2𝑥 + 6𝑦                            ←←← Since 𝑥=𝑟·cos(𝜃) and 𝑦=𝑟·sin(𝜃)

   𝑥² + 𝑦² = 2𝑥 + 6𝑦                     ←←← Since 𝑟²=𝑥²+𝑦²

   𝑥² − 2𝑥 + 𝑦² − 6𝑦 = 0

   𝑥² − 2𝑥 + 1 + 𝑦² − 6𝑦 + 9 = 10 ←←← Complete both squares

   (𝑥 − 1)² + (𝑦 − 3)² = 10            ←←← Since 𝛼²−2𝛼𝛽+𝛽²=(𝛼−𝛽)²

   (𝑥 − 1)² + (𝑦 − 3)² = (√10)²

which is the equation of the circle centred on (1; 3) with radius √10.

   𝑟 = 6·cos(𝜃)

   𝑟² = 6𝑟·cos(𝜃)                       ←←← Multiply everything by 𝑟

   𝑟² = 6𝑥                                 ←←← Since 𝑥=𝑟·cos(𝜃)

   𝑥² + 𝑦² = 6𝑥                           ←←← Since 𝑟²=𝑥²+𝑦²

   𝑥² − 6𝑥 + 𝑦² = 0

   𝑥² − 6𝑥 + 9 + 𝑦² = 9              ←←← Complete the square

   (𝑥 − 3)² + 𝑦² = 9                    ←←← Since 𝛼²−2𝛼𝛽+𝛽²=(𝛼−𝛽)²

   (𝑥 − 3)² + 𝑦² = 3²

which is the equation of the circle centred on (3; 0) with radius 3.

Explicatively,

Dragon.Jade :-)

r = 2cos(θ) + 6sin(θ)

r² = 2rcos(θ) + 6rsin(θ)

x² + y² = 2x + 6y

x² + y² - 2x - 6y = 0