What is 2x^{(2)}+2y^{(2)}+10x-2y-7=0 ?

The solution to 2x^{(2)}+2y^{(2)}+10x-2y-7=0 is Circle with (a,b)=(-5/2 , 1/2),r=sqrt(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^{(2)}+2y^{(2)}+10x-2y-7=0

2 x^{(2)} + 2 y^{(2)} + 10 x - 2 y - 7 = 0

(a, b) = (- \frac{5}{2}, \frac{1}{2}), r = 10

Solution steps

2 x^{2} + 2 y^{2} + 10 x - 2 y - 7 = 0

Rewrite

2 x^{2} + 2 y^{2} + 10 x - 2 y - 7 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - \frac{1}{2})^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, \frac{1}{2}), r = 10

x^{2} + 4 x + y^{2} - 3 y = 0

x^{2} + y^{2} - 4 y - 15 = 45

x^{2} + y^{2} - 4 x + 6 y - 23 = 0

x^{2} + y^{2} - 8 y - 4 = - 16

y^{2} = 8 x - 2 - 13 - x^{2}

What is 2x^{(2)}+2y^{(2)}+10x-2y-7=0 ?
The solution to 2x^{(2)}+2y^{(2)}+10x-2y-7=0 is Circle with (a,b)=(-5/2 , 1/2),r=sqrt(10)