The angle between the tangents drawn from the | vedclass

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

Question

(A) The equation of the circle is $(x-7)^2+(y+1)^2=25$. The center of the circle is $C(7, -1)$ and the radius $r = 5$. The distance $d$ from the origi

Vedclass · Accepted Answer

$2 \arcsin(5/\sqrt{50})$

Vedclass · Answer

(A) The equation of the circle is $(x-7)^2+(y+1)^2=25$. The center of the circle is $C(7, -1)$ and the radius $r = 5$. The distance $d$ from the origin $O(0, 0)$ to the center $C(7, -1)$ is $d = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50}$. Let $	heta$ be the angle between the tangents. The angle between the tangents is given by $2 \alpha$,where $\sin(\alpha) = \frac{r}{d}$. Thus,$\sin(\alpha) = \frac{5}{\sqrt{50}}$. Therefore,the angle between the tangents is $2 \arcsin\left(\frac{5}{\sqrt{50}}ight)$.

The angle between the tangents drawn from the origin to the circle $(x-7)^2+(y+1)^2=25$ is

Similar Questions

The equation of a line passing through $(7, 4)$ and touching the circle $x^2 + y^2 - 6x + 4y - 3 = 0$ is:

$A$ circle touches the $y$-axis at the point $(0,4)$ and passes through the point $(2,0)$. Which of the following lines is not a tangent to this circle?

If the line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = a^2$,then the point of contact is

Let $A$ be the centre of the circle $x^{2}+y^{2}-2x-4y-20=0$. Let $B(1,7)$ and $D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at the point $(1, -2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module