The angle between the two tangents drawn from | vedclass

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

Question

(B) Let $PT$ and $PQ$ be the tangents drawn from point $P(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$,and let $\angle TPQ = 	heta$.If $O$ i

Vedclass · Accepted Answer

$2	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$

Vedclass · Answer

(B) Let $PT$ and $PQ$ be the tangents drawn from point $P(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$,and let $\angle TPQ = 	heta$.If $O$ is the center of the circle,then $\angle TPO = \angle QPO = 	heta/2$.In the right-angled triangle $	riangle OTP$,we have:$	an(	heta/2) = \frac{OT}{PT} = \frac{a}{\sqrt{S_1}}$,where $S_1 = \alpha^{2} + \beta^{2} - a^{2}$.Therefore,$	heta/2 = 	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$.Hence,$	heta = 2	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$.

The angle between the two tangents drawn from the point $(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$ is:

Similar Questions

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y - 4 = 0$ is:

If $y+c=0$ is a tangent to the circle $x^2+y^2-6x-2y+1=0$ at $(a, 4)$,then

The equation of a normal to the circle $x^2+y^2-2x=0$ that is parallel to the line $x+2y-3=0$ is

The straight line $x + 2y = 1$ meets the coordinate axes at $A$ and $B$. $A$ circle is drawn through $A, B$ and the origin. Then the sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is

The equation of the tangent to the curve $x^2+y-7=4x$ at the point $(1,10)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module