What is 4x^{(2)}+40x+4y^{(2)}-32y=5 ?

The solution to 4x^{(2)}+40x+4y^{(2)}-32y=5 is Circle with (a,b)=(-5,4),r= 13/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^{(2)}+40x+4y^{(2)}-32y=5

4 x^{(2)} + 40 x + 4 y^{(2)} - 32 y = 5

(a, b) = (- 5, 4), r = \frac{13}{2}

Solution steps

4 x^{2} + 40 x + 4 y^{2} - 32 y = 5

Rewrite

4 x^{2} + 40 x + 4 y^{2} - 32 y = 5

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 4)^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 5, 4), r = \frac{13}{2}

x^{2} + y^{2} - 10 x + 2 y + 20 = 0

(X + 8)^{2} + (Y + 6)^{2} = 100

(x^{2} + y^{2} - 1)^{2} > 9

9 x^{2} - 6 x + 9 + 9 y^{2} + 24 y + 144 = 19 + 9 + 144

(- 1 - x)^{2} + (2 - y)^{2} = 25

What is 4x^{(2)}+40x+4y^{(2)}-32y=5 ?
The solution to 4x^{(2)}+40x+4y^{(2)}-32y=5 is Circle with (a,b)=(-5,4),r= 13/2