SOLUTION: The point p(3,-4) lies ..........the circle X^2+y^2+6x+8y=0 a) inside b) outside c)on d)none

Question 1083554: The point p(3,-4) lies ..........the circle X^2+y^2+6x+8y=0
a) inside
b) outside
c)on
d)none
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Answer: Choice B) outside

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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 outside of the circle.

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

Visual Proof:

(Image generated by GeoGebra which is free graphing software)
Note: The equation is equivalent to

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