What is 2x^2+2y^2-6x+10y+7=0 ?

The solution to 2x^2+2y^2-6x+10y+7=0 is Circle with (a,b)=(3/2 ,-5/2),r=sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+2y^2-6x+10y+7=0

2 x^{2} + 2 y^{2} - 6 x + 10 y + 7 = 0

(a, b) = (\frac{3}{2}, - \frac{5}{2}), r = 5

Solution steps

2 x^{2} + 2 y^{2} - 6 x + 10 y + 7 = 0

Rewrite

2 x^{2} + 2 y^{2} - 6 x + 10 y + 7 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, - \frac{5}{2}), r = 5

a^{2} + b^{2} = 10

(x - 3)^{2} + (y - 4)^{2} = 9

x^{2} + y^{2} - 6 x + 4 y - 36 = 0

(x - 2)^{2} + (y - 4)^{2} = 100

x^{2} - 6 x + y^{2} - 7 = 0

What is 2x^2+2y^2-6x+10y+7=0 ?
The solution to 2x^2+2y^2-6x+10y+7=0 is Circle with (a,b)=(3/2 ,-5/2),r=sqrt(5)