Two tangents are drawn from a point P to the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the angle between the tangents be $	heta = 	an^{-1}\left(\frac{12}{5}ight)$. Thus,$	an 	heta = \frac{12}{5}$.Since $	an 	heta = \frac{

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(B) Let the angle between the tangents be $	heta = 	an^{-1}\left(\frac{12}{5}ight)$. Thus,$	an 	heta = \frac{12}{5}$.Since $	an 	heta = \frac{12}{5}$,we have $\sin 	heta = \frac{12}{13}$ and $\cos 	heta = \frac{5}{13}$.Let $r$ be the radius of the circle. From the equation $x^2+y^2-2x-4y+4=0$,we have $(x-1)^2 + (y-2)^2 = 1^2$,so $r=1$.Let $\alpha = 	heta/2$ be the angle between the tangent and the line joining the center to the external point $P$. Then $	an \alpha = \frac{r}{PA} = \frac{1}{PA}$.Also,$	an 	heta = \frac{2 	an \alpha}{1 - 	an^2 \alpha} = \frac{2/PA}{1 - 1/PA^2} = \frac{2 PA}{PA^2 - 1} = \frac{12}{5}$.$10 PA = 12 PA^2 - 12 \implies 6 PA^2 - 5 PA - 6 = 0 \implies (2 PA - 3)(3 PA + 2) = 0$. Since $PA > 0$,$PA = 3/2$.Area of $\Delta PAB = \frac{1}{2} (PA)^2 \sin 	heta = \frac{1}{2} \left(\frac{3}{2}ight)^2 \left(\frac{12}{13}ight) = \frac{1}{2} \cdot \frac{9}{4} \cdot \frac{12}{13} = \frac{27}{26}$.Area of $\Delta CAB = \frac{1}{2} r^2 \sin(180^\circ - 	heta) = \frac{1}{2} (1)^2 \sin 	heta = \frac{1}{2} \cdot \frac{12}{13} = \frac{6}{13}$.Ratio = $\frac{	ext{Area}(\Delta PAB)}{	ext{Area}(\Delta CAB)} = \frac{27/26}{6/13} = \frac{27}{26} \cdot \frac{13}{6} = \frac{27}{12} = \frac{9}{4}$.

Similar Questions

Two common tangents to the circle ${x^2} + {y^2} = 2{a^2}$ and the parabola ${y^2} = 8ax$ are

The equations of tangents to the circle $x^2 + y^2 - 22x - 4y + 25 = 0$ which are perpendicular to the line $5x + 12y + 8 = 0$ are

The equation of the tangent to the curve given by $x=3 \cos \theta, y=3 \sin \theta$ at $\theta=\frac{\pi}{4}$ is

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module