SOLUTION: The point (1,2) lies outside the circle x^2+y^2=k if k satisfies the condition? a)k>5 b)k<=5 c)k<5 d)0<k<5

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The point (1,2) lies outside the circle x^2+y^2=k if k satisfies the condition? a)k>5 b)k<=5 c)k<5 d)0<k<5      Log On


   

Question 1089207: The point (1,2) lies outside the circle x^2+y^2=k if k satisfies the condition?
a)k>5
b)k<=5
c)k<5
d)0
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
point (1,2)
x%5E2%2By%5E2=k->1%5E2%2B2%5E2=k->5=k
if k=5,point (1,2) would lie on the circle x%5E2%2By%5E2=5
so, the point (1,2) will lie outside the circle if k%3C5 satisfies the condition?
check:
if k=5

if k%3C5, let say k=3

as you can see, the point (1,2) will lie outside the circle
so, your answer is:
c)k%3C5