Question 1045648
{{{y=(1/3)(x-3)^2 - 3}}}
{{{(x-3)^2 + (y+2)^2 = 1}}} 
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{{{y=(1/3)(x-3)^2 - 3}}} if you set {{{x=0}}}, then {{{y= - 3}}}........substitute it in {{{(x-3)^2 + (y+2)^2 = 1}}} 

{{{(x-3)^2 + (-3+2)^2 = 1 }}}

{{{(x-3)^2 + (-1)^2 = 1 }}}

{{{(x-3)^2 + 1= 1}}} 

{{{(x-3)^2 = 1 -1}}}

{{{(x-3)^2 = 0}}} it will be true if

{{{(x-3)= 0}}} and it will be true if {{{x=3}}}

so, real solution to the system above is: {{{x=3}}}, {{{y=-3}}}

or point ({{{3}}},{{{-3}}})

 

{{{drawing( 600, 600, -7, 7, -7, 7,
circle(3,-3,.1),locate(3,-3,p(3,-3)),
 graph( 600, 600, -7, 7, -7,7, (1/3)(x-3)^2 - 3, sqrt(1-(x-3)^2)-2,-sqrt(1-(x-3)^2)-2)) }}}