[x+y=5quad y[sqrt(2)-x|40|],[2x-3y=4]

Solve for x and ysqrt(2)x-sqrt(3)y=0,sqrt(5)x+sqrt(2)y=0

4x+3y=5, 2x-y=2

The equation of the line that touches the curves y=x|x| and x^2+(y^2-2)^2=4 , where x!=0, is y=4sqrt(5)x+20 (b) y=4sqrt(3)x-12 y=0 (d) y=-4sqrt(5)x-20

If x, y are rational numbers such that (x+y)+(x-2y)sqrt(2)=2x-y+(x-y-1)sqrt(5) then

If x, y are rational numbers such that (x+y)+(x-2y)sqrt(2)=2x-y+(x-y-1)sqrt(5) then

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)

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 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

An equilateral triangle whose two vertices are (-2, 0) and (2, 0) and which lies in the first and second quadrants only is circumscribed by a circle whose equation is : (A) sqrt(3)x^2 + sqrt(3)y^2 - 4x +4 sqrt(3)y = 0 (B) sqrt(3)x^2 + sqrt(3)y^2 - 4x - 4 sqrt(3)y = 0 (C) sqrt(3)x^2 + sqrt(3)y^2 - 4y + 4 sqrt(3)y = 0 (D) sqrt(3)x^2 + sqrt(3)y^2 - 4y - 4 sqrt(3) = 0