What is the radius 4x^2+4y^2-3x+5y-5=0 ?

The radius 4x^2+4y^2-3x+5y-5=0 is r=(sqrt(57))/(4sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius 4x^2+4y^2-3x+5y-5=0

radius

4 x^{2} + 4 y^{2} - 3 x + 5 y - 5 = 0

r = \frac{57}{4 2}

Solution steps

4 x^{2} + 4 y^{2} - 3 x + 5 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} - 3 x + 5 y - 5 = 0

in the form of the standard circle equation

(x - \frac{3}{8})^{2} + (y - (- \frac{5}{8}))^{2} = (\frac{57}{4 2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{8}, - \frac{5}{8}), r = \frac{57}{4 2}

And the radius is:

r = \frac{57}{4 2}

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

x^{2} + y^{2} + 14 x - 18 y + 81 = 0

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

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

(x + 12)^{2} + (y - 9)^{2} = 35

What is the radius 4x^2+4y^2-3x+5y-5=0 ?
The radius 4x^2+4y^2-3x+5y-5=0 is r=(sqrt(57))/(4sqrt(2))