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) Given circle: $x^2+y^2+4x-6y+4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$. Centre $O = (-g, -f) = (-2, 3)$. Radius $r = \sqr

Vedclass · Accepted Answer

$2 	an^{-1} \left(\frac{3}{2}ight)$

Vedclass · Answer

(A) Given circle: $x^2+y^2+4x-6y+4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$. Centre $O = (-g, -f) = (-2, 3)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+9-4} = \sqrt{9} = 3$. Distance from origin $P(0,0)$ to centre $O(-2,3)$ is $d = \sqrt{(-2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$. Let $\alpha$ be the angle between the tangent and the line joining the origin to the centre. In the right-angled triangle formed by the origin,the point of tangency,and the centre,$\sin \alpha = \frac{r}{d} = \frac{3}{\sqrt{13}}$. Then $\cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1-\frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$. Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2}$. The angle between the two tangents is $2\alpha = 2 	an^{-1} \left(\frac{3}{2}ight)$.

The angle between the tangents drawn from the origin to the circle $x^2+y^2+4x-6y+4=0$ is

Similar Questions

Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $PQ=6$ and $PR=5$,then the radius of the circle is

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$,then its radius is

The area of the triangle formed by the tangent at $(3, 4)$ to the circle ${x^2} + {y^2} = 25$ and the coordinate axes is

The angle between the tangents drawn from the point $(2, 2)$ to the circle $x^2 + y^2 + 4x + 4y + c = 0$ is $\cos^{-1}\left(\frac{7}{16}\right)$. If two such circles exist,then the sum of the values of $c$ is:

The equations of the tangents to the circle $x^2+y^2=4$ drawn from the point $(4,0)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module