Question

From a point $P$ on the line $4x-3y=6$ two tangents are drawn to the circle $x^2+y^2-6x-4y+4=0.$ If the angle between these tangents is ${\tan }^{-1}\left(\frac{24}{7}\right)$, then $P=$

• A. $\left(1,\frac{-2}{3}\right)$
• B. $\left(2,\frac{2}{3}\right)$
• C. $\left(-1,\frac{-10}{3}\right)$
• D. $(6,6)$

Answer 1

Answer: (D)

Given equation of circle is
$x^2+y^2-6x-4y+4=0$, whose centre is $C(3,2)$ and radius
$=\sqrt{9+4-4}=3$
Let $P(h,k)$ be a point on the line $4x-3y=6$ from which two tangents are drawn to the circle.

Also, let the angle between these tangents is $2\alpha$, as shown in the figure.
Then, clearly $\angle APC=\angle BPC=\alpha$ and $\angle PAC=\angle PBC=90^{\circ }$
Now, we have $2\alpha ={\tan }^{-1}\frac{24}{7}$
$\Rightarrow \tan 2\alpha =\frac{24}{7}$
$\Rightarrow \frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{24}{7}$
$\Rightarrow 7\tan \alpha =12-12{\tan }^2\alpha$
$\Rightarrow 12{\tan }^2\alpha +7\tan \alpha -12=0$
$\Rightarrow 12{\tan }^2\alpha +16\tan \alpha -9\tan \alpha -12=0$
$\Rightarrow 4\tan \alpha (3\tan \alpha +4)-3(3\tan \alpha +4)=0$
$\Rightarrow (4\tan \alpha -3)(3\tan \alpha +4)=0$
$\Rightarrow \tan \alpha =\frac{3}{4}$
[$\because 3\tan \alpha +4\ne 0,$ as $\alpha$ is an acute angle]

Also, we have,
$\sin \alpha =\frac{CA}{CP}=\frac{3}{\sqrt{(h-3{)}^2+(k-2{)}^2}}$
$\Rightarrow \frac{3}{5}=\frac{3}{\sqrt{(h-3{)}^2+(k-2{)}^2}}$
$\Rightarrow (h-3{)}^2+(k-2{)}^2=25$
$\Rightarrow h^2+9-6h+k^2+4-4k=25$
$\Rightarrow h^2+k^2-6h-4k=12$
Since, $(h,k)$ lies on $4x-3y=6$, therefore $4h-3k=6$
$\to h-\frac{6+3k}{4}$ ...(ii)
On putting the value of $h$ in Eq. (i), we get
${\left(\frac{6+3k}{4}\right)}^2+k^2-6\left(\frac{6+3k}{4}\right)-4k=12$
$\Rightarrow \frac{36+9k^2+36k}{16}+k^2-\frac{(36+18k)}{4}-4k=12$
$\Rightarrow \frac{36+9k^2+36k+16k^2-144-72k-64k}{16}=12$
$\Rightarrow 25k^2-100k-108=192$
$\Rightarrow 25k^2-100k-300=0$
$\Rightarrow k^2-4k-12=0$
$\Rightarrow k^2-6k+2k-12=0$
$\Rightarrow k(k-6)+2(k-6)=0$
$\Rightarrow (k-6)(k+2)=0$
$\Rightarrow k=6$ or $k=-2$
When $k=6,$ then $h=6$
when $k=-2,$ then $h=0$