(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`

(i) 7x^(2) + 16x - 15 = 0 (ii) 5y^(2)+ 8y - 21 = 0

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

(i) x^(2) + 6x + 5 = 0 (ii) y^(2) -10y+21 = 0

Tangents drawn from the point P(1,8) to the circle x^(2)+y^(2)-6x-4y-11=0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (A) x^(2)+y^(2)+4x-6y+19=0 (B) x^(2)+y^(2)-4x-10y+19=0 (B) x^(2)+y^(2)-2x+6y-29=0 (D) x^(2)+y^(2)-6x-4y+19=0

I. x^(2) + 4x = 21 II y^(2) - 6y + 8 = 0

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 parabola with focus (0,0) and directrix x+y=4 is x^(2)+y^(2)-2xy+8x+8y-16=0x^(2)+y^(2)+8x+8y-16=0x^(2)+y^(2)-2xy+8x+8y=0x^(2)-y^(2)+8x+8y-16=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 locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)