What is (x-3)^2+(y+5)^2=16 ?

The solution to (x-3)^2+(y+5)^2=16 is Circle with (a,b)=(3,-5),r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-3)^2+(y+5)^2=16

(x - 3)^{2} + (y + 5)^{2} = 16

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

Solution steps

(x - 3)^{2} + (y + 5)^{2} = 16

Rewrite

(x - 3)^{2} + (y + 5)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

y^{2} - 8 x - 16 = 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{25} = 1

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

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

y^{2} = - 4 x

What is (x-3)^2+(y+5)^2=16 ?
The solution to (x-3)^2+(y+5)^2=16 is Circle with (a,b)=(3,-5),r=4