Thread: Translation of axes
View Single Post
  #2  
Old June 23rd, 2009, 05:23 PM
Soroban Soroban is offline
Super Member

  
Join Date: May 2006
Location: Lexington, MA (USA)
Posts: 7,286
Thanks: 555
Thanked 4,647 Times in 3,710 Posts
Soroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond reputeSoroban has a reputation beyond repute
Default
Hello, mj.alawami!

Quote:
Find the new equation of the circle of the equation x^2+y^2-4x+6y+9\:=\:0
after the translation that moves the origin to the point (2,-3).

Attempt: .\begin{array}{c}x\:=\:x-2 \\ y\:=\:y+3 \end{array}

(x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9\:=\:0

x^2+y^2-8x+12y+48\:=\:0

Am I correct?

Yes! . . . Good work!


\begin{array}{cccccc}\text{The original circle is:} & (x-2)^2 + (y+2)^2 \:=\:4 &\Rightarrow& \text{Center: }(2,\,\text{-}3),\;r = 2 \\ \\[-3mm] \text{The new circle is:} & (x-4)^2 + (y+6)^2 \:=\: 4 &\Rightarrow& \text{Center: }(4,\,\text{-}6),\;r = 2 \end{array}

And this checks out . . .
Reply With Quote