What is (x-(-3))^2+(y-7/13)^2=(1/2)^2 ?

The solution to (x-(-3))^2+(y-7/13)^2=(1/2)^2 is Circle with (a,b)=(-3, 7/13),r= 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-(-3))^2+(y-7/13)^2=(1/2)^2

(x - (- 3))^{2} + (y - \frac{7}{13})^{2} = (\frac{1}{2})^{2}

(a, b) = (- 3, \frac{7}{13}), r = \frac{1}{2}

Solution steps

(x - (- 3))^{2} + (y - \frac{7}{13})^{2} = (\frac{1}{2})^{2}

Rewrite

(x - (- 3))^{2} + (y - \frac{7}{13})^{2} = (\frac{1}{2})^{2}

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{7}{13})^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{7}{13}), r = \frac{1}{2}

x^{2} + y^{2} - 6 x - 8 y - 38 = 0

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

x^{2} - 2 x + y^{2} - 35 = 0

\frac{x ^{2} + y ^{2} - 4 x - 9}{4} = 0

x^{2} + y^{2} - 1.09 x - 7.64 y = 58.18

What is (x-(-3))^2+(y-7/13)^2=(1/2)^2 ?
The solution to (x-(-3))^2+(y-7/13)^2=(1/2)^2 is Circle with (a,b)=(-3, 7/13),r= 1/2