What is (x-4)^2+(y-3)^2=2[(x+2)^2+(y-1)^2] ?

The solution to (x-4)^2+(y-3)^2=2[(x+2)^2+(y-1)^2] is Circle with (a,b)=(-8,-1),r=4sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-4)^2+(y-3)^2=2[(x+2)^2+(y-1)^2]

(x - 4)^{2} + (y - 3)^{2} = 2 [(x + 2)^{2} + (y - 1)^{2}]

(a, b) = (- 8, - 1), r = 45

Solution steps

(x - 4)^{2} + (y - 3)^{2} = 2 [(x + 2)^{2} + (y - 1)^{2}]

Rewrite

(x - 4)^{2} + (y - 3)^{2} = 2 [(x + 2)^{2} + (y - 1)^{2}]

in the form of the standard circle equation

(x - (- 8))^{2} + (y - (- 1))^{2} = (45)^{2}

Therefore the circle properties are:

(a, b) = (- 8, - 1), r = 45

x^{2} + y^{2} + 2 x - 4 y = 11

(x - 37.5)^{2} + (y + 2.5)^{2} = 1^{2}

(y - 2)^{2} + (x - 8)^{2} = 64

x^{2} + y^{2} - 18 x + 4 y + 60 = 0

x^{2} + y^{2} - 10 x - 16 y - 9 = 0

What is (x-4)^2+(y-3)^2=2[(x+2)^2+(y-1)^2] ?
The solution to (x-4)^2+(y-3)^2=2[(x+2)^2+(y-1)^2] is Circle with (a,b)=(-8,-1),r=4sqrt(5)