What is (x-0.5)^2+(y-1)^2=(1/2)^2 ?

The solution to (x-0.5)^2+(y-1)^2=(1/2)^2 is Circle with (a,b)=(0.5,1),r= 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-0.5)^2+(y-1)^2=(1/2)^2

(x - 0.5)^{2} + (y - 1)^{2} = (\frac{1}{2})^{2}

(a, b) = (0.5, 1), r = \frac{1}{2}

Solution steps

(x - 0.5)^{2} + (y - 1)^{2} = (\frac{1}{2})^{2}

Rewrite

(x - 0.5)^{2} + (y - 1)^{2} = (\frac{1}{2})^{2}

in the form of the standard circle equation

(x - 0.5)^{2} + (y - 1)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (0.5, 1), r = \frac{1}{2}

x^{2} + y^{2} + 6 x + 14 y + 37 = 0

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

x^{2} + y^{2} + 10 x + 18 y + 42 = 0

x^{2} + y^{2} + 6 x - 4 y - 15 = 0

(y + 2)^{2} = 20 - (x - 4)^{2}

What is (x-0.5)^2+(y-1)^2=(1/2)^2 ?
The solution to (x-0.5)^2+(y-1)^2=(1/2)^2 is Circle with (a,b)=(0.5,1),r= 1/2