What is (x-3)^2+(y+4)^2=64 ?

The solution to (x-3)^2+(y+4)^2=64 is Circle with (a,b)=(3,-4),r=8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-3)^2+(y+4)^2=64

(x - 3)^{2} + (y + 4)^{2} = 64

(a, b) = (3, - 4), r = 8

Solution steps

(x - 3)^{2} + (y + 4)^{2} = 64

Rewrite

(x - 3)^{2} + (y + 4)^{2} = 64

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 4))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (3, - 4), r = 8

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

x^{2} + y^{2} + 2 x - 3 y - 15 = 0

x^{2} + y^{2} - 22 x + 6 y + 129 = 0

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

2 x^{2} + 2 y^{2} + 44 x + 28 y = 52

What is (x-3)^2+(y+4)^2=64 ?
The solution to (x-3)^2+(y+4)^2=64 is Circle with (a,b)=(3,-4),r=8