Determine whether the given point is inside, on or outside the given circle: (5,3); $\left( x-3\right)^{2} +\left( y-2\right)^{2} =20$.

Hint: In this question it is given that we have to determine whether the given point (5,3) is inside, on or outside the given circle: $\left( x-3\right)^{2} +\left( y-2\right)^{2} =20$. So to solve this we will be using one method, which says that if (a,b) be any point and
If $$\left( a-3\right)^{2} +\left( b-2\right)^{2} < 20$$, then (a,b) lies inside the circle.
If $$\left( a-3\right)^{2} +\left( b-2\right)^{2} = 20$$, then (a,b) lies on the circle.
And If $$\left( a-3\right)^{2} +\left( b-2\right)^{2} > 20$$, then (a,b) lies outside the circle.

Complete step-by-step solution:
The given point is (5,3),
Now let’s just put the point in the left side of the equation,
$\left( x-3\right)^{2} +\left( y-2\right)^{2}$
=$\left( 5-3\right)^{2} +\left( 3-2\right)^{2} $
=$2^{2}+1^{2}$ = 4+1 = 5, which is less than 20.
i.e,$\left( 5-3\right)^{2} +\left( 3-2\right)^{2} $ < 20.
So we can say that the given point is inside the circle.

Note: While solving this type of question you need to keep in mind, in order to find the solution you have to put the points on the left hand side of the given equation. If the left hand side(LHS) value is greater than the right hand side(RHS) then we can say that the points lie outside the circle and if equal to the RHS value then the points lie on the circle and if less than RHS then the point must be inside the circle.