The diagram below is of 3 circles have 3 centres A, B and C and they are collinear.

enter image description here

The equations of the circumferences of the outer circles are

${(x + 12)^2 + (y + 15)^2 = 25}$


${(x - 24)^2 + (y - 12)^2 = 100}$

The question is to find the equation of the centre circle.

I can tell the centres of the 2 circles are (-12, -15) and (24, 12)

Using the distance formula I can tell that the distance between AC is 45 and therefore the radius is 15 by subtracting the other 2 radii.

I am not sure how to get the coordinates of the centre circle. It is not the midpoint.

The gradient is ${3\over 4}$. Should I use this to somehow use the ratio to find the centre of the centre circle?

  • $\begingroup$ center of the first circle is $(-12,-15)$ $\endgroup$ – gt6989b Nov 9 '15 at 22:13
  • $\begingroup$ @gt6989b I was worried about that. I'll delete previous comment. $\endgroup$ – Sam Weatherhog Nov 9 '15 at 22:28

Here is a simple geometric method to find the coordinates of $B$ using similar right triangles (some of the other answers use this but do not explain that they are doing so and skip some steps).

So we know that point $A$ is at $(-12, -15)$ and point $C$ is at $(24,12)$. Also observe that the radius of circle $A$ is $5$ and the radius of circle $C$ is $10$. As you said, the distance is $45$, and so the diameter of circle $B$ must be $30$ (yielding the more important radius of $15$). We then imagine a right triangle with hypotenuse $\overline{AB}$ and a horizontal leg parallel with the $x$-axis (the other leg is obviously parallel with the $y$-axis). We then notice that the two triangles are similar. We thus conclude that $${\;\;\overline{AB}\;\; \above 1pt \;\;\overline{AC}\;\;} = {\;\;\overline{AB}_x\;\; \above 1pt \;\;\overline{AC}_x\;\;} \qquad \text{and} \qquad {\;\;\overline{AB}\;\; \above 1pt \;\;\overline{AC}\;\;} = {\;\;\overline{AB}_y\;\; \above 1pt \;\;\overline{AC}_y\;\;}$$ We can see that $\overline{AB} = 20$, $\overline{AC} = 45$, and $\overline{AC}_x = 36$ from our calculations above, and a little arithmetic yields $\color{red}{\overline{AB}_x = 16}$. Using $\overline{AC}_y = 27$, we then find that $\color{red}{\overline{AB}_y = 12}$.

We get $B_x$ by taking $A_x + \overline{AB}_x = -12 + 6 = \color{red}{4 = B_x}$
We get $B_y$ by taking $A_y + \overline{AB}_y = -15 + 12 = \color{red}{-3 = B_y}$.
Thus, we get our answer by using the coordinates of the center and the radius, yielding
$$\color{red}{(y+3)^2 + (x-4)^2 = 15^2}$$

Here's a picture I threw together in paint to clarify (hopefully... not the greatest drawing!)
enter image description here


$\bar{AB} = 5 + 15 = 20$ and so $B$ is $20/45 = 4/9$ of the way up, so it must be that its coordinates are $B_x = -12 + 36*4/9$ and $B_y = -15 + 27*4/9$.


Hint: We know from the equations that $$ A=(-12,-15)\qquad\text{and}\qquad C=(24,12) $$The distance between $A$ and $C$ is $\sqrt{36^2+27^2}=45$.

The radius of the circle around $A$ is $5$.

The radius of the circle around $C$ is $10$.

That leaves $30$ as the diameter of the circle around $B$.

Therefore, the distance from $A$ to $B$ is $5+15=20$ and the distance from $B$ to $C$ is $10+15=25$.

Thus, $$ \left(C_x-B_x\right)+\left(B_x-A_x\right)=C_x-A_x=36 $$ Furthermore, the distance in $x$ between the centers is in the same ratio as the distance between the centers: $$ \frac{C_x-B_x}{B_x-A_x}=\frac{25}{20} $$ The same is true for the $y$ coordinates: $$ \left(C_y-B_y\right)+\left(B_y-A_y\right)=C_y-A_y=27 $$ and $$ \frac{C_y-B_y}{B_y-A_y}=\frac{25}{20} $$ Now you can compute $B_x$ and $B_y$.

  • 1
    $\begingroup$ how is this a hint if it gives the answer? $\endgroup$ – Brevan Ellefsen Nov 9 '15 at 22:31
  • $\begingroup$ It gives the center and the diameter, which is a lot, but not all. $\endgroup$ – robjohn Nov 9 '15 at 22:33
  • $\begingroup$ @BrevanEllefsen: I've removed the computation of $B$. $\endgroup$ – robjohn Nov 9 '15 at 22:37

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