Let the equation of circle(R) be:(x-a)²+(y-b)²=r²
The points on y=x which are 5√2 away from (0,0)
are (5,5) and (-5,-5). One of this is a point of tangency 
of R with the line y=x.
Finding point of intersection of R with y=x
=>(x-a)²+(x-b)²=r²
The given equation must have only 1 solution
=>(x-(a+b)/2)²+((a²+b²-r²)/2)-((a+b)/2)²=0
=>x=(a+b)/2 and (a²+b²-r²)/2)-((a+b)/2)²=0
=>a+b=2x => a+b=+/- 10 --- (1)
=>r²=(a-b)²/2 --- (2)
length of chord = distance between of points of intersection
of y+x=0 with R. Let these be(x1,-x1) and (x2,-x2)
=> √2(x1-x2)² = 2√22
=> (x1+x2)²-4x1x2=44 --- (3)
x1,x2 are roots of (x-a)²+(x+b)²=(a-b)²/2
up on simplification and then using eq(1) we get
x²+(b-a)+25=0 => x1+x2=a-b and x1x2=25
Substitute in eq(3)
(a-b)²=144 => a-b = +/-12 --- (4)
Solve (1) and (4) using 4 cases we get
(a,b)=(11,-1),(-1,11),(1,-11),(-11,1)
Only (-11,1) satisfy the condition that (-12,2) is inside R
The equation of circle R:
(x+11)²+(y-1)²=12²/2
=>x²+y²+22x-2y+50=0
to find intersection with x=-y², substitute x=-y²
=> we have to solve Quartic eqn
y^4-21y²-2y+50=0
whose solutions are
y= 1.59229,-1.72132, -4.2071
and  4.33612
=> x=-2.53542,-2.962934,-17.69971,-18.80194
The Points of intersection are:
(-2.53542,1.59229)
(-2.962934,-1.72132)
(-17.69971,-4.2071) and
(-18.80194,4.33612)

Edited on July 22, 2008, 11:54 am

Posted by Praneeth on 2008-07-22 11:45:26