, where is the centre and is the radius.
From the information given, we can generate three equations in three unknowns, so that we can solve them simultaneously for and .
Substituting point A into the equation gives
Substituting point B into the equation gives
This circle's derivative is given by:
This is the derivative, which gives us the gradient of the tangent at all points on the circle.
If it's tangent at B is the line , we can rearrange this to read .
So the gradient of the tangent at point B is .
So at point the gradient is .
Substituting these values into the derivative gives
We can now solve for and .
So far we have:
We also know:
Finally, we know that
Now we finally have enough information for the equation of the circle: