# Thread: Coordinate Geometry in the (x,y) plane

1. ## Coordinate Geometry in the (x,y) plane

Hey everyone,

Okay so i'm going through some past papers now as exams are starting to approach and Im struggling with a few question but there is one inperticular thats really confused me. Could you help me please?

The circle C, with centre at the point A, has equation x^2 + y^2 - 10x + 9 = 0.

Find

(a) The coordinates of A

(c) the coordinates of the points at which C crosses the x-axis.

Ive stared and stared at this question and dont truely know where to begin. I know how to find the coordinates of the centre point when its in the form (x-a)^2 + (y-b)^2 = r^2, so my first thought was to arrange this equation into this form, but it seems im having trouble doing this. Im not sure if im thinking in the right direction or not. It could be staring me in the face.

Hope you can help

Steve

2. To put a circle in standard form you need to complete the square

$\displaystyle x^2 + y^2 - 10x + 9 = 0 \iff x^2 - 10x \quad + y^2= -9$

Now take half of the x coefficient and square it and add it to both sides to get

$\displaystyle x^2 - 10x + 5^2+ y^2= -9+5^2$

Now if you factor the x terms it will be a perfect square.

Note it will always factor into $\displaystyle (x + \text{half of the x coefficient})^2$

So we get

$\displaystyle (x-5)^2+y^2=16$

Can you finish from here?

3. Thats brilliant thanks!

So the coordinates would be (5,0)?