(D) Given equation is $x^2+y^2-6x+10y-2=0$. Completing the square for $x$ and $y$ terms: $(x^2-6x+9) + (y^2+10y+25) - 9 - 25 - 2 = 0$. $(x-3)^2 + (y+5

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(D) Given equation is $x^2+y^2-6x+10y-2=0$. Completing the square for $x$ and $y$ terms: $(x^2-6x+9) + (y^2+10y+25) - 9 - 25 - 2 = 0$. $(x-3)^2 + (y+5)^2 = 36$. To transform the origin to $(3, -5)$,we substitute $x = X+3$ and $y = Y-5$,which implies $X = x-3$ and $Y = y+5$. Substituting these into the equation,we get $X^2 + Y^2 = 36$. Thus,the new equation is $x^2+y^2=36$.

Class 11 Mathematics 1 Conic Sections: Circle Equations of circle

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The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ is

AP EAMCET 2021 All AP EAMCET

A
$x^2+y^2=16$
B
$x^2+y^2=9$
C
$x^2+y^2=25$
D
$x^2+y^2=36$

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The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ is

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