A center of a circle with a parabola

[inferi]

hi,

I have this question which is a challenging problem:
in this link you will veiw the graph of the question https://www.physicsforums.com/attachment.php?attachmentid=11650&d=1195680084

the question is to find the center of the circle which it's radius is equal to 1.
the parabola equation is shown in the link and that's all the question.
what i did so far that i took the two points that toch the curve and with the center of the circle and i found the distance between them which is equal to the radius 1.
so let's say that the center of the circle (h,k) and the point that tocuh the curve in the first querter is (x1,y1) and the other point is (-x2,y2) and the x2 is mines because it is in the second querter.
do the final equations are:
(x1-h)^2+(y1-k)^2=(1)^2
(-x2-h)^2+(y2-k)^2=(1)^2

the center of the circle is on the y-axies so h=0 and we substitute x1 and -x2 in the parabola equation so we can get y.

x1^2+(x1^2-k)^2=1
x2^2+(x2^2-k)^2=1

but there are three unknown so i need anothe equation i asked a teacher and he told me to find the conecation between the parabola and the two points and i do not know how to finish this question.
so anyone can please help? thank you

 

[HallsofIvy]

Call the center of the circle $(0, y_1)$ and the point at the circle touches the parabola $(x_0, x_0^2)$ (with $x_0$ positive). The slope of the line from $(0,y_1)$ to $(x_0,x_0^2)$ is $(x_0^2- y_1)/x_0$ while the slope of the tangent line to the parabola is $2x_0$. Since those two lines are perpendicular the product of the two slopes is -1:
$$2x_0\frac{x_0^2- y_1}{x_0}= 2(x_0^2- y_1)= -1.$$

 

[inferi]

Thanks i got it the center is (0,1.25) and i used the other point just to make sure and it gave me the same answer.