What is 4x^2+4y^2-10x-12y=20 ?

The solution to 4x^2+4y^2-10x-12y=20 is Circle with (a,b)=(5/4 , 3/2),r=(sqrt(141))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2+4y^2-10x-12y=20

4 x^{2} + 4 y^{2} - 10 x - 12 y = 20

(a, b) = (\frac{5}{4}, \frac{3}{2}), r = \frac{141}{4}

Solution steps

4 x^{2} + 4 y^{2} - 10 x - 12 y = 20

Rewrite

4 x^{2} + 4 y^{2} - 10 x - 12 y = 20

in the form of the standard circle equation

(x - \frac{5}{4})^{2} + (y - \frac{3}{2})^{2} = (\frac{141}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{4}, \frac{3}{2}), r = \frac{141}{4}

16 x^{2} + 96 x + 16 y^{2} - 40 y - 315 + 315 = 0 + 315

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

x^{2} + (y - 5)^{2} = 100

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

x^{2} - 10 x + y^{2} + 7 y + 12 = 0

What is 4x^2+4y^2-10x-12y=20 ?
The solution to 4x^2+4y^2-10x-12y=20 is Circle with (a,b)=(5/4 , 3/2),r=(sqrt(141))/4