Find the other end of the diameter of the
circle `x^(2) + y^(2) -8x-8y+27=0` if one end
of it is `(2,3)`.

Find the other end of the diameter of the circle x^(2) +y^(2) -8x + 27 =0 if one end of it is (2,3)

(i) Find the other end of the diameter of the circle x^(2)+y^(2)-8x-8y+27=0 if one endof it (2,3). (ii) Snow that A(3,-1) lies on the circle x^(2)+y^(2)-2x+4y=0 . Also find the other end of the diameter through A.

(i) Find the other end of the diameter of the circle x^(2)+y^(2)-8x-8y+27=0 if one endof it (2,3). (ii) Snow that A(3,-1) lies on the circle x^(2)+y^(2)-2x+4y=0 . Also find the other end of the diameter through A.

If one end of a diameter of the circle 2x^(2)+2y^(2)-4x-8y+2=0 is (3,2), the other end is

If one end of diameter of the circle 2x^2+2y^2-4x-8y+2=0 is (3,2) , then the other end is :

If one end of the diameter of the circle x^2 + y^2 - 8x - 4y + c = 0 is (-3, 2), then the other end is :

If one end of the a diameter of the circle 2x^(2)+2y^(2)-4x-8y+2=0 is (3,2), then find the other end of the diameter.

If one end of a diameter of the circle 3x^(2)+3y^(2)-9x+6y+5=0 is (1,2), then the other end is

If the coordinates at one end of a diameter of the circle x^(2) + y^(2) - 8x -4y + c=0 are (11,2) the coordinates of the other end are

If the coordinates of one end of a diameter of the circle x^(2) + y^(2) + 4x-8y + 5=0 is (2,1), the coordinates of ihe other end is