Circle $C_1$ has equation $(x - 13)^2 + (y + 4)^2 = 100$ - Scottish Highers Maths - Question 12 - 2018

To find the circle center coordinates for $C_2$, we start from the equation: $C_2: x^2 + y^2 + 14x - 22y + c = 0$ The center of circle $C_2$ can be derived as: ext{Center } C_2 = (-rac{14}{2}, -rac{-22}{2}) = (-7, 11).
Then, we calculate the distance from the center of $C_1$ to the center of $C_2$ using the distance formula: $d = ext{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2) = ext{sqrt}((-7 - 13)^2 + (11 + 4)^2) = ext{sqrt}((-20)^2 + (15)^2) = ext{sqrt}(400 + 225) = ext{sqrt}(625) = 25.$
Now we know that this distance must equal the radius of $C_1$, which is $10$, plus the radius $r_2$ of $C_2$. Since $C_1$ has a radius of $10$, and $d = r_1 + r_2$, we have:

The distances from $C_1$ to $C_2$ can be calculated as follows:

1. Distance from center of $C_1$ to $P$ (10) and from $C_2$ to $P$ (15).
2. Therefore, the ratio $r_1 : r_2 = 10 : 15 = 2 : 3.$

Using the section formula, the coordinates of point $P$, which divides the line connecting the centers of $C_1$ and $C_2$ in the ratio $2:3$, can be calculated as: P = rac{m(x_2) + n(x_1)}{m+n}, rac{m(y_2) + n(y_1)}{m+n},
Where:

• $m = 2$, $n = 3$
• Center of $C_1 = (13, -4)$
• Center of $C_2 = (-7, 11)$
  Substituting in: P_x = rac{2(-7) + 3(13)}{2 + 3} = rac{-14 + 39}{5} = rac{25}{5} = 5,
  P_y = rac{2(11) + 3(-4)}{5} = rac{22 - 12}{5} = rac{10}{5} = 2.
  Thus, the coordinates of $P$ are $(5, 2)$.

To determine the equation of circle $C_3$, we start with the known center $P(5, 2)$ and the radius, which is $10$ (the radius of $C_1$ since $C_3$ touches $C_1$ internally). The general formula for a circle is: $(x - h)^2 + (y - k)^2 = r^2,$
Where $(h, k)$ are the coordinates of the center and $r$ is the radius. Therefore, substituting the values: $(x - 5)^2 + (y - 2)^2 = 10^2,$
which simplifies to: $(x - 5)^2 + (y - 2)^2 = 100.$
Expanding gives us the final equation: $x^2 - 10x + 25 + y^2 - 4y + 4 = 100,$
Thus, $x^2 + y^2 - 10x - 4y - 71 = 0$