Draw in rectangular coordinate system. It is found that L passes through the focus of C 1, and the circle to be solved is tangent to L, so (0,0) is the point on the circle to be solved. Then the equation of the straight line passing through point (0,0) and perpendicular to L is Y=2X, and the diameter is above it, so the center of the circle is above it. Let the center of the circle be P(X 1, Y 1.

And C 1P=PC2, so [(x1-1) 2+(y1-2) 2]1/2 = (x12+y 2.

So 2 * x 1+4 * y 1 = 5-②.

Simultaneous ① ② Find x1=12, Y 1= 1.

The radius r = (x12+y 2)1/2 = (51/2) * (1/2).

So the circle is: (x-0.5) 2+(y- 1) 2 = 2.5.