What is (x-3)^2+(y-(-2))^2=(1/2)^2 ?

The solution to (x-3)^2+(y-(-2))^2=(1/2)^2 is Circle with (a,b)=(3,-2),r= 1/2

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=(1/2)^2

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

(a, b) = (3, - 2), r = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 2), r = \frac{1}{2}

x^{2} + y^{2} + 2 x + 4 y - 44 = 0

(x - 1)^{2} + (y + 4)^{2} = 4

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

x^{2} + y^{2} - 6 x - 14 y - 111 = 0

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

What is (x-3)^2+(y-(-2))^2=(1/2)^2 ?
The solution to (x-3)^2+(y-(-2))^2=(1/2)^2 is Circle with (a,b)=(3,-2),r= 1/2