Equation of the circle through origin which cuts intercepts of length `a\ a n d\ b` on axes is a. `x^2+y^2=a x+b y=0` b. `x^2+y^2-a x-b y=0` c. `x^2+y^2+b x+a y=0` d. none of these

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 equation of the circle passing through the origin and cutting intercepts a and b on the coordinate axes is x^2 + y^2 - ax - by =0

The equation of the circle passing through the point (1,1) and having two diameters along the pair of lines x^2-y^2-2x+4y-3=0 is a. x^2+y^2-2x-4y+4=0 b. x^2+y^2+2x+4y-4=0 c. x^2+y^2-2x+4y+4=0 d. none of these

The equation of diameter of a circle passing through origin is x+y=1 and the greatest distance of any point of the circle from the diameter is sqrt(5) , then an equation of the circle is: (A) x^2+y^2-2x+4y=0 (B) x^2+y^2-4x+2y=0 (C) x^2+y^2+2x-2y=0 (D) none of these

The equation of the parabola whose vertex is (a ,0) and the directrix has the equation x+y=3a is a. x^2+y^2+2x y+6a x+10 a y+7a^2=0 b. x^2-2x y+y^2-6a x+10 a y+7a^2=0 c. x^2-2x y+y^2+6a x+10 a y+7a^2=0 d.None of these

The equation of the circle passing through (1, 0) and (0, 1) and having smallest possible radius is : (A) 2x^2 + y^2 - 2x-y=0 (B) x^2 + 2y^2 - x - 2y = 0 (C) x^2 = y^2 - x - y = 0 (D) x^2 + y^2 + x + y = 0

2(a x-b y)+(a+4b)=0 , 2(b x+a y)+(b-4a)=0

A circle with center (a , b) passes through the origin. The equation of the tangent to the circle at the origin is (a) a x-b y=0 (b) a x+b y=0 b x-a y=0 (d) b x+a y=0

The equation of a line through (2, -4) which cuts the axes so that the intercepts are equal in magnitude is : (A) x+y+2=0 (B) x-y+2=0 (C) x+y+6=0 (D) x+y-6=0

The equation of straight line passing through (-a ,0) and making a triangle with the axes of area T is (a) 2T x+a^2y+2a T=0 (b) 2T x-a^2y+2a T=0 (c) 2T x-a^2y-2a T=0 (d)none of these