What is 2(x-1)^2+2(y+5)^2=40 ?

The solution to 2(x-1)^2+2(y+5)^2=40 is Circle with (a,b)=(1,-5),r=2sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2(x-1)^2+2(y+5)^2=40

2 (x - 1)^{2} + 2 (y + 5)^{2} = 40

(a, b) = (1, - 5), r = 25

Solution steps

2 (x - 1)^{2} + 2 (y + 5)^{2} = 40

Rewrite

2 (x - 1)^{2} + 2 (y + 5)^{2} = 40

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 5))^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (1, - 5), r = 25

x^{2} + y^{2} - 8 x + 6 y + 9 = 0

(x - 2)^{2} + (y + 4)^{2} = 16

16 - x^{2} - y^{2} = 0

y^{2} = 8 - x^{2}

x^{2} + y^{2} + 8 x - 10 y - 8 = 0

What is 2(x-1)^2+2(y+5)^2=40 ?
The solution to 2(x-1)^2+2(y+5)^2=40 is Circle with (a,b)=(1,-5),r=2sqrt(5)