Question

From a point on the line $4x-3y=6$ tangents are drawn to the circle $x^2+y^2-6x-4y+4=0$ which make an angle of ${\tan }^{-1}\frac{24}{7}$ between them. Find the co-ordinates of all such points and the equations of tangents.

Answer 1

$2\alpha ={\tan }^{-1}\frac{24}{7}\therefore \frac{2t}{1-t^2}=\frac{24}{7}$
or $24t^2+14t-24=0$
or $(8t-6)(3t+4)=0$
$\therefore t=\tan \alpha =\frac{6}{8}=\frac{3}{4}=\frac{3}{PQ}$
$\therefore PQ=4,CQ=3\therefore PC=5$.
If the co-ordinates of $P$ be $(h,k)$, then
$(h-3{)}^2+(k-2{)}^2=25$
The point $(h,k)$ lies on $4x-3y=6$
$\therefore 4h-3k=6\therefore \frac{4h-6}{3}=k.....\left(2\right)$
Putting in $\left(1\right)$, we get
$(h-3{)}^2+{(\frac{4h-6}{3}-2)}^2=25$
or $(h-3{)}^2[1+\frac{16}{9}]=25\therefore (h-3)=\pm 3$
$\therefore h=6$ or $0\therefore k=6$ or $-2$
Hence the points are $(6,6)$ or $(0,-2)$.
The equation of the tangents are given by
$y-6=m(x-6)$ or $mx-y+6-6m=0$
Apply $p=r$ we get $\frac{3m-2+6-6m}{\sqrt{(m^2+1)}}=3$
$\therefore (4-3m{)}^2=9(m^2+1)$
or $0.m^2+24m-7=0$
$\therefore m=\infty$ or $\frac{7}{24}\therefore x-6=0$
or $7x-24y+102=0$
Similarly, we can find the tangents passing through $(0,-2)$ as $\frac{y+2}{x}=\infty$ or $\frac{7}{24}$ i.e. $x=0$ or $7x-24y-48=0$
Note : The tangents from a point can also be obtained by using $S{S}_1=T^2$.