When the origin is shifted to (2,3) then the original equation of `x^(2)+y^(2)+4x+6y+12=0` is

Without change of axes the origin is shifted to (h, k), then from the equation x^(2)+y^(2)-4x+6y-7=0 , then therm containing linear powers are missing, then point (h, k) is

Keeping coordinate axes parallel, the origin is shifted to a point (1, –2), then transformed equation of x^2 + y^2 = 2 is -

When the origin is shifted to a suitable point, the equation 2x^(2)+y^(2)-4x+4y=0 transformed as 2x^(^^)2+y^^2-8x+8y+18=0^(@) The point to which origin was shifted is

Find the point at which origin is shifted such that the transformed equation of x^(2)+2y^(2)-4x+4y-2=0 has no first degree term. Also find the transformed equation .

Without changing the direction of the axes, the origin is transferred to the point (2,3). Then the equation x^(2)+y^(2)-4x-6y+9=0 changes to

The coordinates of the point where origin is shifted is (-1,2) so that the equation 2x^(2)+y^(2)-4x+4y=0 become?

When the origin is shifted to (h,k) ,then the equation (x)/(a)+(y)/(b)=2 is changed to (X)/(a)+(Y)/(b)=0, then (h)/(a)+(k)/(b)=

When the origin is shifted to the point (5,-2) then the transformed equation of the curve xy+2x-5y+11=0 is XY=K then k is

When the origin is shifted to (h,k) ,then the equation (x)/(a)+(y)/(b)=2 is changed to (X)/(a)+(Y)/(b)=0 then (h)/(a)+(k)/(b) = a) 1 b)-1 c)2 d) - 2