How do you find the standard form given $x^2+y^2-6x+4y+14=0$ ?

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Answer 1 · 2018-08-08T08:05:47

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Complete the square on both $x$ and $y$

$x^2 -6x + 9 + y^2 + 4y + 4 = -14 + 9 + 4$

$(x-3)^2 + ( y +2)^2 = -1$

$=> EE x , y in CC$

In other words when you try and sketch thus graph on a cartesian plane, nothing shows up, no real values of $x$ and $y$ satisfy this equation

How do you find the standard form given x^2+y^2-6x+4y+14=0x^2+y^2-6x+4y+14=0?