MATHS QUESTION

Let Point M(x1, y1) be the tangent point of circle on line AB
r be the radius of circle,
and correspondingly, the center of circle have the coordinates C(r, r)

since AB tangent to circle at M
CM ┴ AB
∠AMC = 90deg
∴ ∠AMC = ∠AOB = 90deg
∠MAO = ∠OAB (common angle)
△MAO ~ △OAB (AA similarity)

Since △OAB is a isosceles triangle, △MAO is a isosceles triangle

Similarly, △MBO ~ △OAB

And △MBO = △MAO (AAS) with common side OM

Therefore, MA = MB, and M is the mid-point on AB and have the coordinates (3,3)

OC = √(r² + r²) = (√2)r
OM = √(3² + 3²) = 3√2
However, OM = OC + r

Hence, 3√2 = (√2)r + r = (1 + √2) r
r = 3√2 / (1 + √2)
= 3√2 * (1 - √2) / [(1 + √2) * (1 - √2)]
= (3√2 - 6) / (1 - 2)
= (3√2 - 6) / (-1)
= 6 - 3√2

Finally, the center of circle have the coordinates (6 - 3√2, 6 - 3√2)