What is x^2-8x+y^2-7y+1=0 ?

The solution to x^2-8x+y^2-7y+1=0 is Circle with (a,b)=(4, 7/2),r=(sqrt(109))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2-8x+y^2-7y+1=0

x^{2} - 8 x + y^{2} - 7 y + 1 = 0

(a, b) = (4, \frac{7}{2}), r = \frac{109}{2}

Solution steps

x^{2} - 8 x + y^{2} - 7 y + 1 = 0

Rewrite

x^{2} - 8 x + y^{2} - 7 y + 1 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - \frac{7}{2})^{2} = (\frac{109}{2})^{2}

Therefore the circle properties are:

(a, b) = (4, \frac{7}{2}), r = \frac{109}{2}

\frac{( x - 1 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{4} = 1

(x + 4)^{2} + (y - 4)^{2} = 22

x^{2} + y^{2} - 3 x = 15 - 6 y

2 x^{2} + x + 2 y^{2} = 6

z^{2} = 1 - y^{2}

What is x^2-8x+y^2-7y+1=0 ?
The solution to x^2-8x+y^2-7y+1=0 is Circle with (a,b)=(4, 7/2),r=(sqrt(109))/2