What is (x-4)^2+(y+3)^2=5 ?

The solution to (x-4)^2+(y+3)^2=5 is Circle with (a,b)=(4,-3),r=sqrt(5)

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)^2=5

(x - 4)^{2} + (y + 3)^{2} = 5

(a, b) = (4, - 3), r = 5

Solution steps

(x - 4)^{2} + (y + 3)^{2} = 5

Rewrite

(x - 4)^{2} + (y + 3)^{2} = 5

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 3))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (4, - 3), r = 5

x^{2} + z^{2} = 36

(x + 1.6)^{2} + (y + 9.8)^{2} = 26

(x - 3)^{2} + (y - 1)^{2} = 20

\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{2 ^{2}} = 1

x^{2} + y^{2} + 10 x + 20 y + 77 = 0

What is (x-4)^2+(y+3)^2=5 ?
The solution to (x-4)^2+(y+3)^2=5 is Circle with (a,b)=(4,-3),r=sqrt(5)