If (1,2)(2,a) are extremities of a diameter of the circle `x^(2)+y^(2)-3x-4y+6=0` then a=

One of the diameter of the circle x^(2)+y^(2)-2x+4y-4=0 , is

One of the diameters of the circle x^(2)+y^(2)-12x+4y+6=0 is given by

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

The line x+3y=0 is a diameter of the circle x^(2)+y^(2)-6x+2y=0

If (-1, 3) and (alpha, beta) are the extremities of the diameter of the circle x ^(2) + y ^(2) - 6x + 5y - 7 = 0 , find the coordinates (alpha, beta ) .

The length of the diameter of the circle x^(2)+y^(2)-4x-6y+4=0 is -

The line x+3y=0 is a diameter of the circle x^(2)+y^(2)+6x+2y=0

If (6, –3) is the one extremity of diameter to the circle x^(2) + y^(2) – 3x + 8y – 3 = 0 then its other extremity is

If y = 2x + m is a diameter to the circle x^(2) + y^(2) + 3x + 4y - 1 = 0 , then find m