Question

simplify the equation to x^2+y^2=r^2
x^2+y^2+6x-10y+33

Answer 1

Something is missing!

Answer 2

all of it equals 0

Answer 3

You can write in the form \((x-a)^2+(y-b)^2=r^2\) by completing squares. That's \((x^2+6x+9)(y^2-10y+25=1 \implies (x+3)^2+(x-5)^2=1\).

Answer 4

whew bit it still will not be
\[x^2+y^2=r^2\]
\[(x+3)^2+(y-5)^2=1\]

Answer 5

x^2+6x+9-9+y^2-10y+25-25+33=0
(x+3)^2+(y-5)^2=-33+9+25
(x+3)^2+(y-5)^2=1

Answer 6

Sorry \((x+3)^2+(y-5)^2=1\) :D

Answer 7

The vertex is \((-3,5)\). The x-intercept can be found by plugging \(y=0\) in the equation. Doing so gives \((x+3)^2+25=1 \implies (x+3)^2=-24\), but \((x+3)^2 \ge0\). So, There is no x-intercept.