Respuesta :

(x-3)^2+(y-2)^2=25
plug in those values for x,y to see if they satisfy the equation (=25)

Answer: The point is (D) (3, 0).

Step-by-step explanation:  We are to select the point that does not lie on the circle with centre at (3, 2) and radius 5 units.

We know that the equation of a circle with centre at (g, h) and radius 'r' units is given by

[tex](x-g)^2+(y-h)^2=r^2.[/tex]

If (g, h) = (3, 2) and r = 5 units, then the equation of the circle becomes

[tex](x-3)^2+(y-2)^2=5^2\\\\\Rightarrow (x-3)^2+(y-2)^2=25.[/tex]

If (x, y) = (0, 6), then

L.H.S. = (0-3)² +(6-2)² = 9 + 16 = 25 = R.H.S.

So, (0, 6) lie on the given circle.

If (x, y) = (3, -3), then

L.H.S. = (3-3)² +(-3-2)² = 0 + 25 = 25 = R.H.S.

So, (3, -3) lie on the given circle.

If (x, y) = (-2, 2), then

L.H.S. = (-2-3)² +(2-2)² = 25 + 0 = 25 = R.H.S.

So, (-2, 2) lie on the given circle.

If (x, y) = (3, 0), then

L.H.S. = (3-3)² +(0-2)² = 0 + 4 = 4 ≠ R.H.S.

So, (0, 6) does not lie on the given circle.

Thus, the correct option is (D) (3, 0).