Question 1083554
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Answer: <font color=red>Choice B) outside</font>
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Explanation:


The point (3,-4) is equal to (x,y). 
(x,y) = (3,-4)
The x coordinate is x = 3
The y coordinate is y = -4


Plug x = 3 and y = -4 into the the equation x^2+y^2+6x+8y=0 and simplify.


x^2+y^2+6x+8y=0
(3)^2+(-4)^2+6(3)+8(-4)=0
9+16+6(3)+8(-4)=0
9+16+18-32=0
25+18-32=0
33-32=0
1=0

The last equation (1 = 0) is false.
So the point is NOT on the boundary of the circle.


The left side result is larger than 0, telling us that the point P is going to be on the <font color=red>outside of the circle</font>. 


If we got a negative result on the left side, then P would be on the inside of the circle.


Visual Proof:
<img src = "https://i.imgur.com/GwY0Uoo.png">
(Image generated by <a href = "https://www.geogebra.org/home?ggbLang=en">GeoGebra</a> which is free graphing software)
Note: The equation {{{x^2+y^2+6x+8y=0}}} is equivalent to {{{(x+3)^2+(y+4)^2 = 25}}}
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