What is (x+3)^2+(y-4)^2=7 ?

The solution to (x+3)^2+(y-4)^2=7 is Circle with (a,b)=(-3,4),r=sqrt(7)

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=7

(x + 3)^{2} + (y - 4)^{2} = 7

(a, b) = (- 3, 4), r = 7

Solution steps

(x + 3)^{2} + (y - 4)^{2} = 7

Rewrite

(x + 3)^{2} + (y - 4)^{2} = 7

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 4)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 4), r = 7

(x - 2)^{2} + (y + 2)^{2} = 20

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

(x + 1)^{2} + (y - \frac{2}{3})^{2} = \frac{281}{3}^{2}

x^{2} + y^{2} + 8 y - 128 = 0

2 x^{2} + 2 y^{2} + 8 x - 12 y - 6 = 0

What is (x+3)^2+(y-4)^2=7 ?
The solution to (x+3)^2+(y-4)^2=7 is Circle with (a,b)=(-3,4),r=sqrt(7)