What is (x+4)^2+(y-3/4)^2=125 ?

The solution to (x+4)^2+(y-3/4)^2=125 is Circle with (a,b)=(-4, 3/4),r=sqrt(125)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+4)^2+(y-3/4)^2=125

(x + 4)^{2} + (y - \frac{3}{4})^{2} = 125

(a, b) = (- 4, \frac{3}{4}), r = 125

Solution steps

(x + 4)^{2} + (y - \frac{3}{4})^{2} = 125

Rewrite

(x + 4)^{2} + (y - \frac{3}{4})^{2} = 125

in the form of the standard circle equation

(x - (- 4))^{2} + (y - \frac{3}{4})^{2} = (125)^{2}

Therefore the circle properties are:

(a, b) = (- 4, \frac{3}{4}), r = 125

(x + 6)^{2} + (y - 1)^{2} = 16

(x - 2)^{2} + (y + 5)^{2} = 25

(x - 6)^{2} + y^{2} = 25

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

x^{2} + y^{2} + 9 x = 0

What is (x+4)^2+(y-3/4)^2=125 ?
The solution to (x+4)^2+(y-3/4)^2=125 is Circle with (a,b)=(-4, 3/4),r=sqrt(125)