If the inverse point of the point (-1,1) with respect to the circle x^2+y^2-2 x+2 y-1=0 is (p, q) , then p^2+q^2=

Step-by-step solution

The equation of pole w.r.t. point (-1,1) to the circle x^2+y^2-2 x+2 y-1=0 is aligned & -x+y-(x-1)+(y+1)-1=0 & 2 x-2 y=1 aligned Since, inverse point of (-1,1) is foot of the perpendicular from (-1,1) to the line 2 x-2 y-1=0 gathered (p, q)= ( b^2 x₁+a b y₁-a c (a^2+b^2 ) , b^2 y₁-a b x₁-b c (a^2+b^2 ) ) = ( 4(-1)+(2)(2)(1)-(2)(-1) 8 , 4(1)-(2)(-2)(-1)-(-2)(-1) 8 ) (p, q)= ( 1 4 , -1 4 ) p^2+q^2= 1 16 + 1 16 = 1 8 gathered