What is (x+1/2)^2+(y-5)^2= 25/4 ?

The solution to (x+1/2)^2+(y-5)^2= 25/4 is Circle with (a,b)=(-1/2 ,5),r= 5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+1/2)^2+(y-5)^2= 25/4

(x + \frac{1}{2})^{2} + (y - 5)^{2} = \frac{25}{4}

(a, b) = (- \frac{1}{2}, 5), r = \frac{5}{2}

Solution steps

(x + \frac{1}{2})^{2} + (y - 5)^{2} = \frac{25}{4}

Rewrite

(x + \frac{1}{2})^{2} + (y - 5)^{2} = \frac{25}{4}

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, 5), r = \frac{5}{2}

x^{2} + y^{2} - 4 x + 11 - 84 = 0

2 x^{2} + 2 y^{2} - 12 x + 4 y + 18 = 0

(x + 7)^{2} + y^{2} = 16

(x + 1)^{2} + (y - 1)^{2} = 9 (x - 2)^{2} + 9 (y + 1)^{2}

(x - 9)^{2} + (y - 8)^{2} = 81

What is (x+1/2)^2+(y-5)^2= 25/4 ?
The solution to (x+1/2)^2+(y-5)^2= 25/4 is Circle with (a,b)=(-1/2 ,5),r= 5/2