What is the center x^2-14x+49+y^2+8y+16-25=0 ?

The center x^2-14x+49+y^2+8y+16-25=0 is (7,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2-14x+49+y^2+8y+16-25=0

center

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

(7, - 4)

Solution steps

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

Rewrite

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

And the center is:

(7, - 4)

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

x^{2} + y^{2} + 2 x + 10 y + 18 = 0

x^{2} + (y - 1)^{2} = 9

(x - 11)^{2} + (y + 2)^{2} = 16

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

What is the center x^2-14x+49+y^2+8y+16-25=0 ?
The center x^2-14x+49+y^2+8y+16-25=0 is (7,-4)