Prove that the following pairs of circles intersect orthogonally :
`x^(2) +y^(2) - 2ax + 2by + c = 0 and x^(2) + y^(2) + 2bx + 2ay - c = 0`.

If two circles x^(2) + y^(2) - 2ax + c = =0 and x^(2) + y^(2) - 2by + c = 0 touch each other , then c =

If circles x^(2) + y^(2) + 2gx + 2fy + c = 0 and x^(2) + y^(2) + 2x + 2y + 1 = 0 are orthogonal , then 2g + 2f - c =

Which of the following pairs(s) of curves is/are orthogonal? y^2=4a x ; y=e^(-x/(2a)) y^2 = 4ax; x^2 = 4ay at (0,0) x y=a^2; x^2-y^2=b^2 y=a x ; x^2+y^2=c^2

Equation of the circle on the common chord of the circles x ^(2) + y ^(2) -ax =0 and x ^(2) + y ^(2) - ay =0 as a diameter is

Equation of a circle that cuts the circle x^2 + y^2 + 2gx + 2fy + c = 0 , lines x=g and y=f orthogonally is : (A) x^2 + y^2 - 2gx - 2fy - c = 0 (B) x^2 + y^2 - 2gx - 2fy - 2g^2 - 2f^2 - c =0 (C) x^2 + y^2 + 2gx + 2fy + g^ + f^2 - c = 0 (D) none of these

Two circles x^(2) + y^(2) + ax + ay - 7 = 0 and x^(2) + y^(2) - 10x + 2ay + 1 = 0 will cut orthogonally if the value of a is

Equation of the circle which cuts each of the circles x ^(2) + y ^(2) + 2gx + c =0, x ^(2) + y ^(2) + 2g _(1) x +c =0 and x ^(2) + y^(2) + 2hx + 2ky + a=0 orthogonally is