What is x^2+y^2+10x+16y-107=0 ?

The solution to x^2+y^2+10x+16y-107=0 is Circle with (a,b)=(-5,-8),r=14

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2+10x+16y-107=0

x^{2} + y^{2} + 10 x + 16 y - 107 = 0

(a, b) = (- 5, - 8), r = 14

Solution steps

x^{2} + y^{2} + 10 x + 16 y - 107 = 0

Rewrite

x^{2} + y^{2} + 10 x + 16 y - 107 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 8))^{2} = 1 4^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 8), r = 14

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

(x - 1)^{2} + y^{2} = 6

2 x - 4 y - x^{2} - y^{2} - 1 \geq 0

(x - (- 15))^{2} + (y - 37)^{2} = 2^{2}

(x + 1)^{2} + (y + 3)^{2} = 25

What is x^2+y^2+10x+16y-107=0 ?
The solution to x^2+y^2+10x+16y-107=0 is Circle with (a,b)=(-5,-8),r=14