What is (x-3)^2+(y+2)^2=10 ?

The solution to (x-3)^2+(y+2)^2=10 is Circle with (a,b)=(3,-2),r=sqrt(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

(x - 3)^{2} + (y + 2)^{2} = 10

(a, b) = (3, - 2), r = 10

Solution steps

(x - 3)^{2} + (y + 2)^{2} = 10

Rewrite

(x - 3)^{2} + (y + 2)^{2} = 10

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 2))^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (3, - 2), r = 10

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

(x - 2)^{2} + (y + 4)^{2} = 25

x = x^{2} + y^{2}

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y = - 32

What is (x-3)^2+(y+2)^2=10 ?
The solution to (x-3)^2+(y+2)^2=10 is Circle with (a,b)=(3,-2),r=sqrt(10)