Q) Fill in the blanks. Let $C_1$ be the circle with center $(−8, 0)$ and radius $6$. Let $C_2$ be the circle with center $(8,0)$ and radius $2$. Given a point $P$ outside both circles, let $L_i(P)$ be the length of a tangent segment from $P$ to circle $C_i$. The locus of all points $P$ such that $L_1(P) = 3 L_2(P)$ is a circle with radius _____ and center at _____,_____o--->

A) I found this question intriguing while solving the CMI past papers. Here is the work that I've done to attempt to fill in the blanks however the Complete equation of the Locus of the Circle is missing and I was unable to derive the complete equation.

Firstly, I Decided to take the Locus as a moving point $(x,y)$ and then I expressed, using Pythagoras, the Length of the tangents $L_1$ and $L_2$ when $y=0$.

This Results in, $$(x+8)^2 - 6^2 = 9((x-8)^2 -2^2)$$

Further Implying, $$8x^2 - 160x +512 = 0$$ Giving $x=4$ and $16$, when $y=0$.

Hence the midpoint of $4$ and $16$ is $10$, which would be its center, Thus, giving Center $(10,0)$ and Radius $6$.

Although it may not be simple to use this elementary method to find the radius and the center of a circle all the time. Due to symmetry this question wasn't a challenge However Finding the general equation of the circle eludes me when the case isn't so simple. Any Help in Finding the General equation of a Locus would be Appreciated. Thank you! :)

  • 1
    $\begingroup$ Indeed the fact that the ratio of radii $6/2=3$ is equal to the ratio $3=L_1(P)/L_2(P)$ proves that it is a circle which is identical to a certain "circle of Apollonius" associated with points $-8$ and $8$. Can we reformulate your question in the following way: "can the locus be still a circle is the ratios are not the same ?" $\endgroup$
    – Jean Marie
    Feb 10, 2022 at 22:42
  • $\begingroup$ Yes, I am unclear about that as well. And How would you find the general equation of this circle as I read the solution else where and they attempt to do the following; Using the distance formula and the Pythagorean theorem to get $y^2 +(x+8)^2 −6^2 = 9(y^2 +(x−8)^2 −4)$. Simplifying gives $y^2 +(x−10)^2 = 6^2$. I didnot understand this particular method. $\endgroup$
    – Devansh
    Feb 10, 2022 at 22:46

2 Answers 2


Let us consider the general case with the following notations : $(C_1),(C_2)$ for the circles, $C_1=(a_1,0), \ C_2=(a_2,0)$ for their centers and $r_1,r_2$ for their radii, resp. Let $P=(x,y)$. Let $T_1$ and $T_2$ be two points of tangency (of one of the tangents issued from $P$) onto circles $(C_1),(C_2)$.

enter image description here

Fig. 1: Case $a_1=-4, r_1=2, a_2=1, r_2=1, k=2$. The locus is the red circle.

$$L_1(P)=kL_2(P) \ \ \text{that can be written by squaring:} \ \ (PT_1)^2=k^2 (PT_2)^2\tag{1}$$

Applying Pythagoras theorem in triangles $PT_1C_1$ and $PT_2C_2$ resp.:

$$PC_1^2=PT_1^2+r_1^2 \ \ \text{and} \ \ PC_2^2=PT_2^2+r_2^2$$

As a consequence:

$$PT_1^2=PC_1^2-r_1^2 \ \ \text{and} \ \ PT_2^2=PC_2^2-r_2^2 $$

$$PT_1^2=((x-a_1)^2+y^2)-r_1^2 \ \ \text{and} \ \ PT_2^2=((x-a_2)^2+y^2)-r_2^2 $$

Plugging these expressions into (1) gives:


that can be simplified into:


which, by division by $(k^2-1)$ gives indeed the equation of a circle,

$$(x^2+y^2)-2\underbrace{\frac{k^2a_2-a_1}{k^2-1}}_a x+\underbrace{\frac{(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)}{k^2-1}}_c=0\tag{2}$$

The center of this circle is therefore :

$$\left(\frac{k^2a_2-a_1}{k^2-1},0 \right)$$

which is a weighted combination of the coordinates of the initial centers.

In order to determine its radius, we have to refer to the classical equation of a circle:

$$(x-a)^2+(y-b)^2=R^2 \ \ \iff \ \ x^2+y^2-2ax-2by+\underbrace{a^2+b^2-R^2}_c=0\tag{3}$$

Therefore $R$ is given by

$$R^2=a^2+b^2-c=\left(\frac{k^2a_2-a_1}{k^2-1}\right)^2- \frac{(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)}{k^2-1}$$

The RHS must fulfill a positivity constraint, otherwise the locus, instead of being a circle will be empty... This the case for example when $a_1=-1,a_2=1, r_1=4, r_2=1/2, k=3$.

Remark: This exercice generalizes the "circles of Apollonius"

  • $\begingroup$ Any comment ?... $\endgroup$
    – Jean Marie
    Feb 11, 2022 at 23:31


  • $(x+8)^2 + y^2 > 36$.
  • $(x-8)^2 + y^2 > 36$.

First of all, you can assume that without loss of generality,
$|y| \leq 6.$

That is, $|y| > 6$ implies that the two constraints above are automatically satisfied, regardless of the value of $x$.

Further, the constraint $\left[f(x)\right]^2 > A^2$ translates to
$f(x) < -|A| ~~\text{or}~~ f(x) > |A|.$

Therefore, the two constraints translate to

$$(x + 8) < -\sqrt{36 - y^2} ~~~~\text{or}~~~~ (x + 8) > \sqrt{36 - y^2}.\tag1 $$


$$(x - 8) < -\sqrt{36 - y^2} ~~~~\text{or}~~~~ (x - 8) > \sqrt{36 - y^2}.\tag2 $$

I considered trying to collapse the two constraints into something like

$(x + 8) < -\sqrt{36 - y^2} ~~~~\text{or}~~~~ (x - 8) > \sqrt{36 - y^2}.$

Then, I considered, suppose that
$(x - 8) < -\sqrt{36 - y^2} \leq (x + 8)$.

What can be said (for example) about the comparison of

$$(x + 8) ~~~~\text{against}~~~~\sqrt{36 - y^2}.\tag3 $$

It is true that $(x+8) - (x-8) = 16$
while $\displaystyle ~\sqrt{36 - y^2} - \left[-\sqrt{36 - y^2}\right] \leq 12.$

However, the comparison in (3) above is (still) too vulnerable to various values of $x,y$.

Therefore, I will stay with the constraints in (1) and (2) above.


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