What is the center x^2-7x+y^2+8y-14=0 ?

The center x^2-7x+y^2+8y-14=0 is (7/2 ,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2-7x+y^2+8y-14=0

center

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

(\frac{7}{2}, - 4)

Solution steps

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

Rewrite

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

in the form of the standard circle equation

(x - \frac{7}{2})^{2} + (y - (- 4))^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{7}{2}, - 4), r = \frac{13}{2}

And the center is:

(\frac{7}{2}, - 4)

3 (x - 5)^{2} + 3 (y + 9)^{2} = 48

x^{2} + y^{2} + 8 y = 32

x^{2} + y^{2} - 6 x + 8 y - 11 = 0

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

x^{2} + y^{2} - 12 x + 4 y = - 2

What is the center x^2-7x+y^2+8y-14=0 ?
The center x^2-7x+y^2+8y-14=0 is (7/2 ,-4)