What is (X-4)^2+(Y+3)^2=1 ?

The solution to (X-4)^2+(Y+3)^2=1 is Circle with (a,b)=(4,-3),r=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(X-4)^2+(Y+3)^2=1

(X - 4)^{2} + (Y + 3)^{2} = 1

(a, b) = (4, - 3), r = 1

Solution steps

(X - 4)^{2} + (Y + 3)^{2} = 1

Rewrite

(X - 4)^{2} + (Y + 3)^{2} = 1

in the form of the standard circle equation

(X - 4)^{2} + (Y - (- 3))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (4, - 3), r = 1

\frac{( x - 9 ) ^{2}}{2} + \frac{( y - 5 ) ^{2}}{2} = 1

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

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

y^{2} = 36 - x^{2}

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = \frac{64}{25}

What is (X-4)^2+(Y+3)^2=1 ?
The solution to (X-4)^2+(Y+3)^2=1 is Circle with (a,b)=(4,-3),r=1