What is x^2+y^2-14x+8y+40=0 ?

The solution to x^2+y^2-14x+8y+40=0 is Circle with (a,b)=(7,-4),r=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2-14x+8y+40=0

x^{2} + y^{2} - 14 x + 8 y + 40 = 0

(a, b) = (7, - 4), r = 5

Solution steps

x^{2} + y^{2} - 14 x + 8 y + 40 = 0

Rewrite

x^{2} + y^{2} - 14 x + 8 y + 40 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

x^{2} + y^{2} + 10 x - 8 y + 37 = 0

(x - 5)^{2} + (y + 6)^{2} = 20

\frac{x ^{2}}{16} + \frac{y ^{2}}{16} = 1

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

x^{2} + y^{2} - 10 x + 2 y + 22 = 0

What is x^2+y^2-14x+8y+40=0 ?
The solution to x^2+y^2-14x+8y+40=0 is Circle with (a,b)=(7,-4),r=5