What is ((x-4)^2)/2+((y-3)^2)/2 =1 ?

The solution to ((x-4)^2)/2+((y-3)^2)/2 =1 is Circle with (a,b)=(4,3),r=sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-4)^2)/2+((y-3)^2)/2 =1

\frac{( x - 4 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{2} = 1

(a, b) = (4, 3), r = 2

Solution steps

\frac{( x - 4 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{2} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{2} = 1

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 3), r = 2

3 y + 6 x - 3 x^{2} - 3 y^{2} = - 1

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

x^{2} + y^{2} - 12 x + 20 y + 15 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{49} = 1

x^{2} + y^{2} + 4 x - 6 y - 10 = 0

What is ((x-4)^2)/2+((y-3)^2)/2 =1 ?
The solution to ((x-4)^2)/2+((y-3)^2)/2 =1 is Circle with (a,b)=(4,3),r=sqrt(2)