What is y^2+x^2-18x+12y+101=0 ?

The solution to y^2+x^2-18x+12y+101=0 is Circle with (a,b)=(9,-6),r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^2+x^2-18x+12y+101=0

y^{2} + x^{2} - 18 x + 12 y + 101 = 0

(a, b) = (9, - 6), r = 4

Solution steps

y^{2} + x^{2} - 18 x + 12 y + 101 = 0

Rewrite

y^{2} + x^{2} - 18 x + 12 y + 101 = 0

in the form of the standard circle equation

(x - 9)^{2} + (y - (- 6))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (9, - 6), r = 4

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

x^{2} + (y + 4)^{2} = \frac{1}{9}

x^{2} + (y - 6)^{2} = 9

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

x^{2} - 2 x + 1 + y^{2} + y + \frac{1}{2} = \frac{11}{4}

What is y^2+x^2-18x+12y+101=0 ?
The solution to y^2+x^2-18x+12y+101=0 is Circle with (a,b)=(9,-6),r=4