Translation of axes

mj.alawami · #1 $Old$ June 23rd, 2009, 01:03 PM

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
x=x-2
y=y+3

$(x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9=0$
$x^2+y^2-8x+12y+48=0$

Am I correct?

Soroban · #2 $Old$ June 23rd, 2009, 05:23 PM

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 . . .