What is (x-2)^2+(y-(-4))^2=1.7^2 ?

The solution to (x-2)^2+(y-(-4))^2=1.7^2 is Circle with (a,b)=(2,-4),r=1.7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-2)^2+(y-(-4))^2=1.7^2

(x - 2)^{2} + (y - (- 4))^{2} = 1. 7^{2}

(a, b) = (2, - 4), r = 1.7

Solution steps

(x - 2)^{2} + (y - (- 4))^{2} = 1. 7^{2}

Rewrite

(x - 2)^{2} + (y - (- 4))^{2} = 1. 7^{2}

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = 1. 7^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 1.7

x^{2} + y^{2} + y - 11 = 0

x^{2} - 154 x + y^{2} - 160 y = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{16} = 10

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

9 - x^{2} > y^{2}

What is (x-2)^2+(y-(-4))^2=1.7^2 ?
The solution to (x-2)^2+(y-(-4))^2=1.7^2 is Circle with (a,b)=(2,-4),r=1.7