Question

Transform $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$

Answer 1

Let the co-ordinates $(x,y)$ of any point on the given curve change to $(X,Y)$ on shifting the origin to the point $(3,-5)$, the new axes remaining parallel to the original axes.Then, we get

$x=X+3$ and $y=Y-5$

The given equation is $x^2+y^2-6x+10y-2=0$

Substituting for $x=X+3$ and $y=Y-5$ we get

${(X+3)}^2+{(Y-5)}^2-6(X+3)+10(Y-5)-2=0$

$\Rightarrow X^2+9+6X+Y^2-10Y+25-6X-18+10Y-50-2=0$

$\Rightarrow X^2+Y^2=36$