Because c is tangent to y=x, |a-b|/ (radical number 2)=2* radical number 2①.

C passes through the origin again, so a2-b2=2* root number 2②.

Because the point is in the second quadrant, the solution is a=-2 and b=2.

So the equation of C is (x+2)2+(y-2)2=8.

2) According to the meaning of the question, the distance from one point on the ellipse to two focal points | pf1| pf2 | =10 = 2a.

So a=5, so the standard equation of ellipse is X2/25+Y2/9= 1.

Let Q(x, y) and f (4 4,0)

Therefore |QF|=((x-4)2+y2) radical number, |OF|=4.

Therefore, in order to meet the meaning of the question, we must satisfy QF=OF, that is, ((x-4)2+y2) radical number =4.

That is, (x+2)2+y2= 16 and (x+2)2+(y-2)2=8.

Simultaneous x+y=0, substitute into the equation of circle to get x=-4, y=4(y=-4 rounded).

So there is Q (-4,4). . . . . .

It turns out that it is too difficult to type these things with the keyboard. ...