What is x^2+y^2+5x-10y-5=0 ?

The solution to x^2+y^2+5x-10y-5=0 is Circle with (a,b)=(-5/2 ,5),r=(sqrt(145))/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+5x-10y-5=0

x^{2} + y^{2} + 5 x - 10 y - 5 = 0

(a, b) = (- \frac{5}{2}, 5), r = \frac{145}{2}

Solution steps

x^{2} + y^{2} + 5 x - 10 y - 5 = 0

Rewrite

x^{2} + y^{2} + 5 x - 10 y - 5 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - 5)^{2} = (\frac{145}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 5), r = \frac{145}{2}

a^{2} - 28 + b^{2} = 7

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1

(x - 6)^{2} + (y - 4)^{2} = 8^{2}

(x - 4)^{2} + (k (x) - 3)^{2} = 5^{2}

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

What is x^2+y^2+5x-10y-5=0 ?
The solution to x^2+y^2+5x-10y-5=0 is Circle with (a,b)=(-5/2 ,5),r=(sqrt(145))/2