What is x^2+y^2-8x+4y-29=0 ?

The solution to x^2+y^2-8x+4y-29=0 is Circle with (a,b)=(4,-2),r=7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2-8x+4y-29=0

x^{2} + y^{2} - 8 x + 4 y - 29 = 0

(a, b) = (4, - 2), r = 7

Solution steps

x^{2} + y^{2} - 8 x + 4 y - 29 = 0

Rewrite

x^{2} + y^{2} - 8 x + 4 y - 29 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, - 2), r = 7

x^{2} + y^{2} - 2 x - 8 y + 13 = 0

x^{2} + y^{2} + 16 x - 16 y + 79 = 0

x^{2} + y^{2} - 10 x + 4 y + 4 = 0

\frac{( y - 2 ) ^{2}}{9} + \frac{( x - \frac{1}{2} ) ^{2}}{9} = 1

x^{2} + y^{2} - 4 x + 10 y - 7 = 0

What is x^2+y^2-8x+4y-29=0 ?
The solution to x^2+y^2-8x+4y-29=0 is Circle with (a,b)=(4,-2),r=7