General equation for a circle

[The Builder]

....... this is going to sound daft, but since I can't draw you a diagram.......

Say the wall is the y-axis, and the floor is the x-axis............ roll a perfect circle right into the angle, so that the wall and the floor are both tangents.

Can anyone tell me what the General equation for that circle would be?

[Prudie]

If you take the exact perpendiculars (right angles) from where the carpet just touches each wall and draw them back until they meet, that will be the centre of the circle. Both distances shoud be the same and that will be the radius r and the centre at (r,r).
Equation should be (x-r)^2 + (y-r)^2 =r^2

[ChuckFickens]

Surely it's going to depend on how far you want the tangent points to be away from the corner, you could roll a 1inch circle in there or a 6 foot circle and the wall and floor would always be tangents to where they touched the circle.

(or have a misunderstood something)

[nightmare]

a circle of radius r, centred on the origin, has equation
x^2 + y^2 = r^2.
Since your circle is touching the axes, its centre is now at the point with co-ordinates (r,r).
Therefore, its equation is (x-r)^2 + (y-r)^2 = r^2

[The Builder]

Chucko ........ it's a general equation I'm looking for, so it would cover any r (radius)

Pruders and Nightmare......... thank you very much. I thought it was going to be that. I just wondered if, when that expression is expanded, it might become simpler :o(

So, for a radius of 1, I would have to do it the long way. To explain, I just want to arrive at an expression for a value of y, for whatever value of x that I give it.

[jake-the-peg]

x²+y²=r²
where r is the radius

so

y=squareroot(r²-x²)

Thing is you have to remember that the squareroot can be + or - and so every time you put in an x and geta y you have to assume you are getting 2 ys a positive and a negative

[The Builder]

Jake, I think that equation only relates to a circle of origin (0,0)

[jake-the-peg]

Yup you just offset to move it about

so substitute a and b for x and y a=x-3, b=y+2

then

b=squareroot(r²-a²)

[jake-the-peg]

so as almost said above

(x-a)² + (y-b)² =r²

so y= +/-sqrt(r²-(x-a)²)+a

if you want a definative equation

[The Builder]

Thanks, Jake :o)

[jake-the-peg]

garrrh!##

y= +/-sqrt(r²-(x-a)²)+b

+/- of course means plus and minus

[The Builder]

Many thanks again, to all

[vascop]

Jakes solution is not what you were asking for. You want the special case where a=b=r, in other words nightmare and prudie's solutions.

[vascop]

In other words
y= +/-sqrt(r²-(x-r)²)+r