Question

Let C be the circle (x-1)^2 + y-2)^2 =144 and P be point P(10,10). Which of the following is true?

Answer 1

First, an introduction:  If the equation of the circle were x^2 + y^2 = 144, then the center would be at (0,0) and the radius would be 12.  Note that the distance from the center to P(10,10) is 10sqrt(2), or 14.14.  Thus, in this example, P would be OUTSIDE the circle (since 14.14 is greater than the radius 12).

Now let's focus on (x-1)^2 + (y-2)^2 =144.   Let x = 12 as an example; find the corresponding y:  9^2 + (y-2)^2 = 144, or    (y-2)^2 = 63, and so y-2 is approx. -8 or +8.  Then y (for x = 12) is either approx. -10 or 6:  (12,-10) or (12,6).  Are these inside the circle or outside?

A better way to address this would be as follows:

Find the distance from the center (1, 2) to the point P(10,10).  If this distance is less than 12, the point P is inside C; if greater than 12, P is outside C.

This distance is sqrt( (10-2)^2 + (10-1)^2 ), or sqrt (64+81) = sqrt(145).
This is LARGER than sqrt(144).  Thus, P is OUTSIDE the circle C.