An integer is called a perfect square if it is square of another integer. The number of perfect square points, i.e. $$x$$ and $$y$$ are both perfect squares such that $$(x,y)$$ lies exactly within the circle
$$(x - 12)^2 + (y - 10)^2 = 64$$, is

The given equation is $$(x - 12)^2 + (y - 10)^2 = 64$$

$$(x - 12)^2 + (y - 10)^2 = 8^2$$

It is an equation of a circle with a radius of 8 units & its centre is at (12,10).

It implies that the extreme coordinate of 'x' is = 12+8 = 20.

Similarly, the extreme coordinate of 'y' is = 10+8 = 18.

The square numbers which are less than 20 & 18 are 4, 9 & 16.

For a point (x,y) to lie inside the circle, $$(x - 12)^2 + (y - 10)^2 = 64$$, the value of $$(x - 12)^2 + (y - 10)^2$$ should be less than 64.

Case (i):- (4,4)

$$(4-12)^2+(4-10)^2=8^2+6^2=100 > 64$$. 

Hence, the point lies outside the circle.

As $$(4-12)^2=64$$, all the square points with '4' as x coordinate i.e (4,9), (4,16),  lie outside the circle.

Case (ii):- (9,9)

$$(9-12)^2+(9-10)^2=3^2+1^2=10 < 64$$.

Hence, the point lies inside the circle.

Case (iii):- (16,16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (iv):- (9,4)

$$(9-12)^2+(4-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (v):- (9,16)

$$(9-12)^2+(16-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (vi):- (16, 4)

$$(16-12)^2+(4-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (vii):- (16, 16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

A total of 6 perfect square points lie inside the circle.

Option 'C' is the answer.