What is (x-8)^2+(y-5)^2= 16/9 ?

The solution to (x-8)^2+(y-5)^2= 16/9 is Circle with (a,b)=(8,5),r= 4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-8)^2+(y-5)^2= 16/9

(x - 8)^{2} + (y - 5)^{2} = \frac{16}{9}

(a, b) = (8, 5), r = \frac{4}{3}

Solution steps

(x - 8)^{2} + (y - 5)^{2} = \frac{16}{9}

Rewrite

(x - 8)^{2} + (y - 5)^{2} = \frac{16}{9}

in the form of the standard circle equation

(x - 8)^{2} + (y - 5)^{2} = (\frac{4}{3})^{2}

Therefore the circle properties are:

(a, b) = (8, 5), r = \frac{4}{3}

x^{(2)} + y^{(2)} - 5 x + 4 y = 49

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

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

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

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

What is (x-8)^2+(y-5)^2= 16/9 ?
The solution to (x-8)^2+(y-5)^2= 16/9 is Circle with (a,b)=(8,5),r= 4/3