What is x^2+y^2-7x-3y-51=0 ?

The solution to x^2+y^2-7x-3y-51=0 is Circle with (a,b)=(7/2 , 3/2),r=sqrt(131/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2-7x-3y-51=0

x^{2} + y^{2} - 7 x - 3 y - 51 = 0

(a, b) = (\frac{7}{2}, \frac{3}{2}), r = \frac{131}{2}

Solution steps

x^{2} + y^{2} - 7 x - 3 y - 51 = 0

Rewrite

x^{2} + y^{2} - 7 x - 3 y - 51 = 0

in the form of the standard circle equation

(x - \frac{7}{2})^{2} + (y - \frac{3}{2})^{2} = (\frac{131}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{7}{2}, \frac{3}{2}), r = \frac{131}{2}

x^{2} + 6 x + y^{2} - 12 y - 99 = 0

(x^{2}) + (y^{2}) = 49

8 x + x^{(2)} - 2 y = 64 - y^{(2)}

x^{2} + y^{2} + 4 y = 4

4 x^{2} + 36 x + 4 y^{2} - 8 y + 68 = x^{2} - 6 x + y^{2} - 4 y + 13

What is x^2+y^2-7x-3y-51=0 ?
The solution to x^2+y^2-7x-3y-51=0 is Circle with (a,b)=(7/2 , 3/2),r=sqrt(131/2)