SOLUTION: Find the equation of the locus of a point which moves so that its distance from the point C (3, 4) is always equal to 5. a. x2 + y2 + 6x - 8y = 0 ​ b. x2 + y2 + 6x + 8y = 0

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Question 1146433: Find the equation of the locus of a point which moves so that its distance from the point C (3, 4) is always equal to 5.
a. x2 + y2 + 6x - 8y = 0 ​
b. x2 + y2 + 6x + 8y = 0 ​ ​
​c. x2 + y2 - 6x - 8y = 0
d. x2 + y2 - 6x + 8y = 0
Found 2 solutions by richwmiller, KMST:
Answer by richwmiller(17219) About Me  (Show Source):
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
it has to be the equation of a circle.
For a circle of radius K=any positive number,
centered at (3,4), the equation would be
%28x-3%29%5E2%2B%28y-4%29%5E2=K%5E2 , which "simplifies" to
x%5E3%2By%5E2-6x-8y=K%5E2-3%5E2-4%5E2
The answer is obviously
​c. x2 + y2 - 6x - 8y = 0,
unless they were tricking you,
and there is no positive number K such that K%5E2-3%5E2-4%5E2 is 0 .
(5%5E2-3%5E2-4%5E2 is zero).