Find center of circle with 2 internally touching circles

Question

A third circle is drawn such that:

• both $C_1$ and $C_2$ touch internally
• The centres of $C_1$, $C_2$ and $C_3$ are collinear.

Determine the equation of $C_3$

Circle C1 has the equation $x^2 + y^2 + 6x + 10y + 9 = 0$

$\therefore$ centre $C_1$ is(-3, -5) with a radius of 5

The equation of $C_2$ is $(x-9)^2 + (y-11)^2 = 225$

Centre of $C_2 = (9, 11)$ with a radius of 15

The distance between $r_1$ and $r_2$ is 20

Looking at the answer to this question, it states:

find ratio in which centre of C3 divides line joining centres of C1 and C2 = 3:1

I do not understand how this was obtained. Why is it 3:1 and then how is the centre (6, 7) obtained?

How should I have known that the ratio is 3:1?

The distance between centre $C_1$ x to $C_2$ x is 12, how do I get 6 from this.

The distance between centre $C_1$ y to $C_2$ y is 16 how do I get 7 from ratio.

Answer 1

(a) The radius of C3 is the sum of the radii of C1 and C2. (b) The ratio that C3's center position will divide the line between C1's and C2's center positions, will be exactly the inverse of the ratios of their radii.

Answer 2

The ratio between R1 and R2 is 3:1 so this would give you a hint, (6,7) would be the 3:1 "balanced" centre of C1 and C2 knowing that C1(-3,-5) and C2(9,11): 6=(3*9+(-3))/4 and 7=(3*11+(-5))/4

C3 is in the segment [C1C2] and verify : 2R3 = 20 , one may define A and B points of intersection of C3 with C1 and C2 respectively and continue the exercise

Answer 3

The diameter of $C_1$ is $10$ and the diameter of $C_2$ is $30$, so the diameter of $C_3$ is $40$ and its radius $20$

On the shared diameters: $C_1$ and $C_2$ touch at $(0,-1)$, while $C_1$ and $C_3$ touch at $(-6,-9)$ and $C_2$ and $C_3$ touch at $(18,23)$, so the centre of $C_3$ is $(6,7)$

You can use these to find the ratio and equation