Parametric eqn of curves

Circle 1... (x-3)^2 + (y+4)^2 = 5
Circle 2... (x-4)^2 + (y+2)^2 =1/4
Show that the centre of circle 2 lies on the circumference of circle 1.

Complete the square for both x and y to show the foll 2 expression represent circles. Specify their centres and radii.
x^2-2x+y^2+4y-4=0
2x^2+2x+2y^2-6y+3=0

The general equation of a circle in 2-D Cartesian coordinates is:
(x - a)² + (y - b)² = r²
where (a,b) is the center and r is the radius.

So:
Circle 2... (x-4)^2 + (y+2)^2 = 1/4 has its center at (4, -2)

Plug the coordinates of the center of Circle 2 into the equation for Circle 1. If the equation remains balanced, then the centre of circle 2 lies on the circumference of circle 1. Hint: it is true.



Complete the square for both x and y to show the foll 2 expression represent circles. Specify their centres and radii. I'll do the first one for you:

x^2-2x+y^2+4y-4=0
Group into x and y terms:
(x^2-2x) + (y^2+4y-4) = 0

Now add sufficient constant terms to each section to make perfect squares, and add the same amounts to the other side. First the x terms:
(x^2 - 2x + 1) + (y^2 + 4y - 4) = 0 + 1
(x - 1)^2 + (y^2 + 4y - 4) = 1

Then the y terms:
(x - 1)^2 + (y^2 + 4y + 4) = 1 + 8
(x - 1)^2 + (y + 2)^2 = 9

So, center is (1,-2) and the radius is 3.