The equation of a circle with center at (a,b) and radius r is: (x-a)^2+(y-b)^2=r^2.
This circle has been translated from (0,0) to (a,b).
The equation of the circle in our case can be rewritten as: (x-2)^2+(y-2)^2=2^2
Moving the graph 3 units to the right is: (x-2-3)^2+(y-2)^2=2^2.
Moving the graph 1 unit down is: (x-2-3)^2+(y-2-1)^2=2^2.
This gives the equation: (x-5)^2+(y-3)^2=2^2
So, the center of the circle is: (5,3)
Answer: The new co-ordinates of the center are (5, -2).
Step-by-step explanation: The given equation of a circle is
[tex]x^2-4x+y^2+2y=4~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We are to find the center of the above circle if the circle (i) is translated 3 units to the right and 1 unit down.
The standard equation of a circle with radius r units and center at (h, k) is given by
[tex](x-h)^2+(y-k)^2=r^2.[/tex]
From equation (i), we have
[tex]x^2-4x+y^2+2y=4\\\\\Rightarrow (x^2-4x+4)+(y^2+2y+1)-4-1=4\\\\\Rightarrow (x-2)^2+(y+1)^2=9\\\\\Rightarrow (x-2)^2+(y+1)^2=3^2.[/tex]
Comparing the above equation with the standard equation of a circle, we get
center, (h, k) = (2, -1) and radius, r = 3 units.
If the circle is translated 3 units to the right and 1 unit down, then the co-ordinates of the new center will be
(h', k') = (h+3, k-1) = (2+3, -1-1) = (5, -2).
Thus, the new co-ordinates of the center are (5, -2).
