Determine whether the point (5,2) lies inside or outside the circle whose equation is x^2+y^2 = 30.

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Equatio f the given circle is

`x^2+y^2=30`

`x^2+y^2=(sqrt(30))^2`

cetre of circle (0,0)

radius of circle= sqrt(30)

distance d between (0,0) and point (5,2)

=`sqrt((5-0)^2+(2-0)^2)`

`=sqrt(25+4)`

`=sqrt(29)`

since d is less than radius of the circle So point will inside the circle.

The equation of the circle is `x^2+y^2=30` .

Then the inequality for the interior (the disk) is `x^2+y^2<30` and the inequality for the exterior is `x^2+y^2>30` .

Substitute x=5 and y=2 to get 29<30.

The point lies in the interior.

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