# a circle has an equation 2(x-2)^2+2y=2 find the intercepts, center and the radius show on graph must show work

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Expert Answers

mlehuzzah | Certified Educator

Your equation is not a circle, it is a parabola.

`2(x-2)^2+2y=2`

`(x-2)^2+y=1`

`y=1 - (x-2)^2`

This is a parabola, that opens down, and has it's vertex at (2,1)

The graph is:

However, the following is a circle:

`2(x-2)^2+2y^2=2`

`(x-2)^2+y^2=1`

Any time you see the following:

`(x-h)^2+(y-k)^2=r^2` you have a circle, with center at (h,k) and radius r

Our equation can be rewritten as:

`(x-2)^2+(y-0)^2=1^2`

So this is a circle with center (2,0) and radius 1

Intercepts occur where x=0 or y=0

If y=0, then:

`(x-2)^2+y^2=1`

`(x-2)^2=1`

`(x-2) = +-1`

`x-2 = 1` OR `x-2 = -1`

`x=3` OR `x=1`

Thus we have intercepts at (3,0), and (1,0)

If x = 0, then:

`(0-2)^2+y^2=1`

`4+y^2=1`

`y^2=-3`

This can't happen, so we don't have any other intercepts.

The graph is: