What is (x-5)^2+(y+4)^2=sqrt(100) ?

The solution to (x-5)^2+(y+4)^2=sqrt(100) is Circle with (a,b)=(5,-4),r=sqrt(\sqrt{100)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-5)^2+(y+4)^2=sqrt(100)

(x - 5)^{2} + (y + 4)^{2} = 100

(a, b) = (5, - 4), r = 100

Solution steps

(x - 5)^{2} + (y + 4)^{2} = 100

Rewrite

(x - 5)^{2} + (y + 4)^{2} = 100

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 4))^{2} = (100)^{2}

Therefore the circle properties are:

(a, b) = (5, - 4), r = 100

x^{2} + y^{2} \leq \frac{3}{4}

x^{2} - 8 x + 16 + y^{2} + 2 y + 1 = 4

x^{2} + y^{2} + 3 x - 10 y - 5 = 0

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

x^{2} + y^{2} - 16 y + 39 = 0

What is (x-5)^2+(y+4)^2=sqrt(100) ?
The solution to (x-5)^2+(y+4)^2=sqrt(100) is Circle with (a,b)=(5,-4),r=sqrt(\sqrt{100)}