What is the center (x-3)^2+(y-5)^2=121 ?

The center (x-3)^2+(y-5)^2=121 is (3,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center (x-3)^2+(y-5)^2=121

center

(x - 3)^{2} + (y - 5)^{2} = 121

(3, 5)

Solution steps

(x - 3)^{2} + (y - 5)^{2} = 121

Rewrite

(x - 3)^{2} + (y - 5)^{2} = 121

in the form of the standard circle equation

(x - 3)^{2} + (y - 5)^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (3, 5), r = 11

And the center is:

(3, 5)

(x + 4)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} - 4 x + 6 y - 36 = 0

3 x^{2} + 3 y^{2} - 12 x + 18 y + 3 = 0

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

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

What is the center (x-3)^2+(y-5)^2=121 ?
The center (x-3)^2+(y-5)^2=121 is (3,5)