C1 Circles help please

Given your

$(x-r)^2 + (y-r)^2 = r^2$

What is the centre of that circle ... how far away from the co-ordinate axes is that ... how does that show that it touches the axes?

Note that a circle touches a line just when there is only one point of intersection between them, i.e. when the simultaneous equations for the circle and the line have exactly one solution.

Well the centre of the circle is then r,r with a radius of r. But I don't get why it shows it :/

So..x² + y² -2rx -2ry + r² = 0 if you substitute x = 0 (the y axis) then you get 0 + y² - 0 - 2ry + r² = 0. So y² - 2ry + r² = 0. Then b²-4ac = 0?

really?

how far from the x-axis is the centre, how far is the y-axis from the centre

try a sketch

Very good, by showing the discriminant is zero you've shown that there's only one solution, and hence the circle touches the y axis.

Now do exactly the same with y = 0, and you're done.

post 4

The second question

Sketch a circle wholly in the first quadrant

Lets say a circle with centre (6,9) ... it does not reach the axes so what can you say about the length of r?

ok so r is less than 6 in my example

now use your question to find the centre of the circle ... it will give you an expression using c for r so you can apply the same understanding to that



wholly in the first quadrant means it is all in the first quadrant

So you know that the radius has to be <5 (same as the e.g. I gave)

And you know that r^2 = 74-c

And, I presume, you know that r^2 must be +ve

Yes I know that, but how is it linked to 49<c<74?

As I know r^2 must be +ve as all ^2 numbers are, but then I don't know what step to take :/

The information that you have leads directly to that inequality