Calc III Curve of Intersection Question

I've been staring at this problem for hours, but can't figure out how to begin. I understand the projection of the intersection on the xy-plane is a circle with center (-9/2,7/2), which means the equation of that intersection would start out with (x+9/2)^2+(y-7/2)^2=, but how do i find the radius of this circle? If I can just figure out how to find the radius and plug that into the circle's equation i'll be set.

Consider the paraboloid z = x^2 + y^2. The plane 9 x - 7 y + z - 10 = 0 cuts the paraboloid, its intersection being a curve.
Find "the natural" parametrization of this curve.
Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x coordinate. Using that as your starting point, give the parametrization of the curve on the surface.

Let

z_1 (x,y) = x^2 + y^2
z_2 (x,y) = 10-9x+7y.

At a point (x,y,z) of the intersection, you need z = z_1(x,y) = z_2(x,y) or

x^2 + y^2 = 10-9x+7y
(x+9/2)^2 + (y-7/2)^2 = 10+81/4+49/4 = 265/2

So then
x  =  \sqrt{265/2}  cos t - 9/2
y = \sqrt{265/2} sin t + 7/2
z = whatever x^2 + y^2 works out to.

thank you for your help. I'm still a little lost as to how you went from

x^2 + y^2 = 10-9x+7y

to

(x+9/2)^2 + (y-7/2)^2 = 10+81/4+49/4 = 265/2

what tells you to substitute 9/2 for x and  7/2 for y in the second part to get 81/4 and 49/4? I know I probably shouldn't be asking such a basic question in calc III, but it's been about 3 years since i took calc II (yay!)

>>3
That's just completing the square for y and x.

>>1

That's not actually calculus.