Equation of the circle passing through `A(1,2), B(5,2)` such that the angle subtended by AB at points the circle is `pi/4` is (A) `x^2+y^2-6x-8=0` (B) `x^2+y^2-6x-8y+17=0` (C) `x^2+y^2 -6x+8=0` (D) `x^2+y^2-6x+8y-25=0`

The equation of the circle passing through (0,0) and cutting orthogonally the circles x^(2) + y^(2) + 6x - 15 = 0 , x^(2) + y^(2) - 8y + 10 = 0 is

The equation of the circle which inscribes a suqre whose two diagonally opposite vertices are (4, 2) and (2, 6) respectively is : (A) x^2 + y^2 + 4x - 6y + 10 = 0 (B) x^2 + y^2 - 6x - 8y + 20 = 0 (C) x^2 + y^2 - 6x + 8y + 25 = 0 (D) x^2 + y^2 + 6x + 8y + 15 = 0

The equation of the circle which inscribes a suqre whose two diagonally opposite vertices are (4, 2) and (2, 6) respectively is : (A) x^2 + y^2 + 4x - 6y + 10 = 0 (B) x^2 + y^2 - 6x - 8y + 20 = 0 (C) x^2 + y^2 - 6x + 8y + 25 = 0 (D) x^2 + y^2 + 6x + 8y + 15 = 0

The equation of circle passing through (1,-3) and the points common to the two circt x^2 + y^2 - 6x + 8y - 16 = 0, x^2 + y^2 + 4x - 2y - 8 = 0 is

The circles x^(2)+y^(2)-6x-8y=0 and x^(2)+y^(2)-6x+8=0 are

The equation of the circle passing through the origin which cuts of intercept of length 6 and 8 from the axes is a. x^2+y^2-12 x-16 y=0 b. x^2+y^2+12 x+16 y=0 c. x^2+y^2+6x+8y=0 d. x^2+y^2-6x-8y=0

Show that the circle touch each other: x^2+y^2-6x-2y+1=0 and x^2+y^2-2x-8y+13=0

(3,-4) is the point to which the origin is shifted and the transformed equation is X^(2)+Y^(2)=4 then the original equation is (A) x^(2)+y^(2)+6x+8y+21=0 (B) x^(2)+y^(2)+6x+8y-21,=0 (C) x^(2)+y^(2)-6x+8y+21=0 (D) x^(2)+y^(2)-6x-8y+21=0