The circle k has equation $(x - 4)^2 + (y + 2)^2 = 169$ - Leaving Cert Mathematics - Question 2 - 2022

To determine if the point (11, 10) is on the circle, inside, or outside, we will substitute this point into the equation of the circle.
We calculate the distance from the center (4, -2) to the point (11, 10):
$d = ext{sqrt}((11 - 4)^2 + (10 + 2)^2) = ext{sqrt}((7)^2 + (12)^2) = ext{sqrt}(49 + 144) = ext{sqrt}(193).$

• Since the radius is 13, we compare:
• $ext{sqrt}(193) ext{ vs } 13$.
• Squaring both sides:
  $193 > 169$ which indicates that the point (11, 10) is outside the circle k.

Hence, the answer is: (11, 10) is outside k.

As the circle s has its center at (22, 13) and it touches circle t at (12, 11),

• the radius can be calculated as the distance from the center to (12, 11).
• The radius is computed as follows:
  $r_s = ext{sqrt}((22 - 12)^2 + (13 - 11)^2) = ext{sqrt}(10^2 + 2^2) = ext{sqrt}(100 + 4) = ext{sqrt}(104).$
• Another point on circle s can thus be calculated by adding and subtracting the radius along the coordinate axes.
  For simplicity, one valid point might be (32, 15).

Given that the radius of circle t is half that of s, we have:

• Radius of circle t:
  r_t = rac{r_s}{2} = rac{ ext{sqrt}(104)}{2} = ext{sqrt}(26).
• The center of circle t lies directly between (12, 11) and (22, 13), which is the midpoint.
• The midpoint can be calculated as:
  M = rac{(x_1 + x_2)}{2}, rac{(y_1 + y_2)}{2}
• Applying this:
  M = rac{(22 + 12)}{2}, rac{(13 + 11)}{2} = (17, 12).
  Thus, the coordinates of the center of circle t are (17, 12).