Results 1 to 4 of 4

- Oct 9th 2009, 10:07 AM #1

- Joined
- Oct 2007
- Posts
- 188

## Bit of help please - Cir. Eq.

**Can someone check and help my methodology and confirm my answer.**

Ok, so the following equation is a circle - and im trying to find the centre and radius

**First Consider**

**Compare with**

**we can match**

**with LHS by putting**

that is

**Putting values into RHS, obtain completed square form**

**Now**

**with y in place of x**

**We can match****with RHS by putting**

**that is**

**Putting values into RHS we obtain completed square form**

**Hence centre (10,4) and radius -80**

Thanks for help

- Oct 9th 2009, 11:24 AM #2

- Oct 9th 2009, 12:46 PM #3

- Joined
- Jan 2008
- From
- Big Stone Gap, Virginia
- Posts
- 2,550
- Thanks
- 15
- Awards
- 1

Hi ADY,

Your solution (however wrong) seems to be a bit convoluted.

Look at it simply.

First, group your x and y, and move +36 to the right side.

The trick here is to add a constant to the x and y group that will make the resulting trinomial a perfect square.

How do we do that?

Take half of the coefficient of the linear term, square it, then add it to the group. Do that for both the x and y group. And since we added a certain amount to the left side, we must add that same amount to the right side.

So, here we go.

Half of -10 is -5. We square that to get 25 and we add that to the x group to get .

Half of 8 is 4. We square that to get 16 and we add that to the y group to get .

We added a total of 25 + 16 on the left, so we add that amount to the right to keep things balanced.

Finally, grouping the two perfect square trinomials, we arrive at

Center (5, -4), Radius =

- Oct 9th 2009, 12:49 PM #4

- Joined
- Oct 2007
- Posts
- 188