The equation of the circle which has normals `(x-1).(y-2)=0` and a tangent `3x+4y=6` is (a) `x^2+y^2-2x-4y+4=0` (b) `x^2+y^2-2x-4y+5=0` (c) `x^2+y^2=5` (d) `(x-3)^2+(y-4)^2=5`

Suppose a x+b y+c=0 , where a ,ba n dc are in A P be normal to a family of circles. The equation of the circle of the family intersecting the circle x^2+y^2-4x-4y-1=0 orthogonally is (a) x^2+y^2-2x+4y-3=0 (b) x^2+y^2-2x+4y+3=0 (c) x^2+y^2+2x+4y+3=0 (d) x^2+y^2+2x-4y+3=0

The equation of a circle which has a tangent 3x + 4y = 6 and two normals given by (x-1) (y-2) = 0 is

The ellipse 4x^2+9y^2=36 and the hyperbola a^2x^2-y^2=4 intersect at right angles. Then the equation of the circle through the points of intersection of two conics is (a) x^2+y^2=5 (b) sqrt(5)(x^2+y^2)-3x-4y=0 (c) sqrt(5)(x^2+y^2)+3x+4y=0 (d) x^2+y^2=25

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is (a) x^2+y^2+4x+4y-8=0 (b) x^2+y^2-3x+4y+8=0 (c) x^2+y^2+x+y=0 (d) x^2+y^2-3x-3y-8=0

Find the equation of the circle which cuts the three circles x^2+y^2-3x-6y+14=0,x^2+y^2-x-4y+8=0, and x^2+y^2+2x-6y+9=0 orthogonally.

From the points (3, 4), chords are drawn to the circle x^2+y^2-4x=0 . The locus of the midpoints of the chords is (a) x^2+y^2-5x-4y+6=0 (b) x^2+y^2+5x-4y+6=0 (c) x^2+y^2-5x+4y+6=0 (d) x^2+y^2-5x-4y-6=0

The locus of the image of the point (2,3) in the line (x-2y+3)+lambda(2x-3y+4)=0 is (lambda in R) (a) x^2+y^2-3x-4y-4=0 (b) 2x^2+3y^2+2x+4y-7=0 (c) x^2+y^2-2x-4y+4=0 (d) none of these

Two circle x^2+y^2=6 and x^2+y^2-6x+8=0 are given. Then the equation of the circle through their points of intersection and the point (1, 1) is x^2+y^2-6x+4=0 x^2+y^2-3x+1=0 x^2+y^2-4y+2=0 none of these

The equation of the circel whose diameter is the common chord of the circles x ^(2) + y^(2) +2x +3y+2=0 and x ^(2) + y^(2) +2x -3y-4=0 is-

Prove that the equation of the parabola whose focus is (0, 0) and tangent at the vertex is x-y+1 = 0 is x^2 + y^2 + 2xy - 4x + 4y - 4=0 .