What is 4x^2+4y^2+6x+8y-5=0 ?

The solution to 4x^2+4y^2+6x+8y-5=0 is Circle with (a,b)=(-3/4 ,-1),r=(3sqrt(5))/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+6x+8y-5=0

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

(a, b) = (- \frac{3}{4}, - 1), r = \frac{3 5}{4}

Solution steps

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, - 1), r = \frac{3 5}{4}

(x - 1)^{2} + (y - 2)^{2} = 10

x^{2} + y^{2} - 14 x + 38 = 0

x^{2} + y^{2} + 4 x - 5 = 0

x^{2} - 4 x + 6 y + y^{2} = 4

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

What is 4x^2+4y^2+6x+8y-5=0 ?
The solution to 4x^2+4y^2+6x+8y-5=0 is Circle with (a,b)=(-3/4 ,-1),r=(3sqrt(5))/4