What is the radius (x+7/2)^2+(y-1/2)^2=sqrt(5) ?

The radius (x+7/2)^2+(y-1/2)^2=sqrt(5) is r=sqrt(\sqrt{5)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius (x+7/2)^2+(y-1/2)^2=sqrt(5)

radius

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

r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, \frac{1}{2}), r = 5

And the radius is:

r = 5

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

\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}

What is the radius (x+7/2)^2+(y-1/2)^2=sqrt(5) ?
The radius (x+7/2)^2+(y-1/2)^2=sqrt(5) is r=sqrt(\sqrt{5)}