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! 🙂
Best Answer
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 $(C_1),(C_2)$.
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:
$$((x-a_1)^2+y^2)-r_1^2=k^2[((x-a_2)^2+y^2)-r_2^2]$$
that can be simplified into:
$$(k^2-1)(x^2+y^2)-2(k^2a_2-a_1)x+(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)=0$$
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"