Question 1089207
given:
point ({{{1}}},{{{2}}})
{{{x^2+y^2=k}}}->{{{1^2+2^2=k}}}->{{{5=k}}}

if {{{k=5}}},point ({{{1}}},{{{2}}}) would lie on the circle {{{x^2+y^2=5}}}

so, the point ({{{1}}},{{{2}}}) will lie {{{outside}}} the circle if {{{k<5}}} satisfies the condition?

check:
if {{{k=5}}}
{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,2,.12),locate(1,2,p(1,2)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(-x^2+5),-sqrt(-x^2+5))) }}} 

if {{{k<5}}}, let say {{{k=3}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,2,.12),locate(1,2,p(1,2)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(-x^2+3),-sqrt(-x^2+3))) }}} 

as you can see, the point ({{{1}}},{{{2}}}) will lie {{{outside}}} the circle

so, your answer is:

c){{{k<5}}}