What is (x+3/2)^2+y^2> 7/4+(3/2)^2 ?

The solution to (x+3/2)^2+y^2> 7/4+(3/2)^2 is Circle with (a,b)=(-3/2 ,0),r=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+3/2)^2+y^2> 7/4+(3/2)^2

(x + \frac{3}{2})^{2} + y^{2} > \frac{7}{4} + (\frac{3}{2})^{2}

(a, b) = (- \frac{3}{2}, 0), r = 2

Solution steps

(x + \frac{3}{2})^{2} + y^{2} > \frac{7}{4} + (\frac{3}{2})^{2}

Rewrite

(x + \frac{3}{2})^{2} + y^{2} > \frac{7}{4} + (\frac{3}{2})^{2}

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 0), r = 2

x^{2} + (y - 6)^{2} = 144

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

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

x^{2} + y^{2} - 6 x + 8 y = 75

3 x^{2} + 3 y^{2} + 12 x + 12 = 18 y

What is (x+3/2)^2+y^2> 7/4+(3/2)^2 ?
The solution to (x+3/2)^2+y^2> 7/4+(3/2)^2 is Circle with (a,b)=(-3/2 ,0),r=2