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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 7)^{(2)} + (y - 7)^{(2)} = \frac{( 4 )}{( 9 )}

(x - 2)^{2} + (y - 4)^{2} = 49

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

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

36 x^{2} - 36 + 36 y^{2} + 72 y = 279

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