Two equal circle touch each other externally at C and AB is a common tangent to the circles. Then ?ACB =?

For two equal circles touching externally at C with common tangent AB, the radii to the point of tangency are perpendicular to the tangent. This creates two radii OA and OB perpendicular to AB. Since O, C, and the point of tangency are collinear, triangle AOB is a right triangle with OA and OB as perpendicular legs, making angle ACB = 90°. Option D is correct.