Question

Shew that if the length of the tangent from a point $P$ to the circle $x^2+y^2=a^2$ be four times the length of the tangent from it to the circle $(x-a{)}^2+y^2=a^2$, then $P$ lies on the circle $15x^2+15y^2-32ax+a^2=0$.
Prove also that these three circles pass through two points and that the distance between the centres of the first and third circles is sixteen times the distance between the centres of the second and third circles.

Answer 1

The two circles are given as
$x^2+y^2=a^2.....1$
and ${(x-a)}^2+y^2=a^2$ or $x^2+y^2-2ax=\mathrm{0.....2}$
If $P$ be any point $(h,k)$ satisfying the given condition, $P{T}_1$ and $P{T}_2$ be the lengths of tangents from P upon $1$ and $2$
$P{T}_1=\sqrt{h^2+k^2-a^2}$
$P{T}_2=\sqrt{h^2+k^2-2ah}$
$P{T}_1=4P{T}_2$
$P{T}_1^2=16P{T}_2^2$
Putting the value $h^2+k^2-a^2=16(h^2+k^2-2ah)$
Locus is;-
$15x^2+15y^2-32ax+a^2=\mathrm{0.......3}$
which is the required locus
Again, common chord of $1$ and $2$ is
$2ax=a^2$ or $2x=a....4$
Common chord of $1$ and $3$ is obtained by multiplying $1$ by $15$ and subtracting $3$ from it ; so we get
$32ax-a^2=15a^2$
$2x=a.....5$
$4$ and $5$ are the same.
Hence, the $3$ circles have the same common chord meaning thereby that the three circles pass through $2$ common points.
Again center of $1$ is $(0,0)$ and center of $2$ is $(a,0)$ and center of $3$ is $(\frac{16}{15}a,0)$
$C_1{C}_3=\frac{16a}{15}-0=\frac{16}{15}a$
$C_1{C}_2=a$
$C_2{C}_3=-\frac{16a}{15}-a=\frac{a}{15}$
Therefore, $C_1{C}_3=16\times C_2{C}_3$