Given the hypotenuse AB, point A (-2,0) and point B (4 4,0) of the right triangle ABC, find the trajectory equation of point C.

C is a right angle, so point C is on a circle with AB as its diameter.

(-2+4)/2= 1.

The midpoint coordinate of AB is (1, 0), AB = 6, then the radius =3.

Therefore, equation c is (x-1) 2+y 2 = 9. C can't coincide with a and b, so x doesn't =-2 and 4.

It is known that point A is any point on the circle C: (x-a)2+(y+2)2=a2+ 10, and the symmetrical point of point A about the straight line L: X+2Y+ 1 = 0 is also on the circle. Find the value of number A.

Let the symmetry point of a about a straight line be A', then the perpendicular line of AA' is a straight line L.

Then I must pass through the center of the circle.

The coordinates of the center of the circle are (a, -2). If you substitute l, you will get:

a+2 *(2)+ 1 = 0

That is, a=3.