If theta is the angle between the tangents dr | vedclass

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

Question

(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.So,the center $O$ i

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{5}{12}ight)$

Vedclass · Answer

(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.So,the center $O$ is $(3,-2)$ and the radius $r$ is $1$.The distance $d$ from the point $A(2,3)$ to the center $O(3,-2)$ is $d = \sqrt{(3-2)^2+(-2-3)^2} = \sqrt{1^2+(-5)^2} = \sqrt{26}$.Let $\alpha$ be the half-angle between the tangents. In the right-angled triangle formed by the point $A$,the center $O$,and the point of tangency $P$,we have $\sin(\alpha) = \frac{r}{d} = \frac{1}{\sqrt{26}}$.Then $\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1-\frac{1}{26}} = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}}$.Thus,$	an(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{1/\sqrt{26}}{5/\sqrt{26}} = \frac{1}{5}$.The total angle between the tangents is $	heta = 2\alpha$.Using the formula $	an(2\alpha) = \frac{2	an(\alpha)}{1-	an^2(\alpha)}$,we get $	an(	heta) = \frac{2(1/5)}{1-(1/5)^2} = \frac{2/5}{24/25} = \frac{2}{5} 	imes \frac{25}{24} = \frac{5}{12}$.Therefore,$	heta = 	an^{-1}\left(\frac{5}{12}ight)$.

If $\theta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6x+4y+12=0$,then $\theta=$

Similar Questions

If $x-2y=0$ is a tangent drawn at a point $P$ on the circle $x^2+y^2-6x+2y+c=0$,then the distance of the point $(6,3)$ from $P$ is

If the tangent drawn at the point $P$ on the circle $x^2+y^2+6x+6y=2$ meets the straight line $5x-2y+6=0$ at a point $Q$ on the $Y$-axis,then the length of $PQ$ is

If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$,then $m_{1}-m_{2}$ is equal to

The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

The angle between the tangents to the circle $x^2 + y^2 = 169$ at the points $(5, 12)$ and $(12, -5)$ is ............. $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module