The angle between a pair of tangents drawn fr | vedclass

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

Question

(D) The equation of the given circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$.Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get

Vedclass · Accepted Answer

$x^{2}+y^{2}+4x-6y+9=0$

Vedclass · Answer

(D) The equation of the given circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$.Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get center $A(-2, 3)$ and radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{4+9-(9 \sin^{2} \alpha + 13 \cos^{2} \alpha)} = \sqrt{13 - 9 \sin^{2} \alpha - 13 \cos^{2} \alpha} = \sqrt{13(1-\cos^{2} \alpha) - 9 \sin^{2} \alpha} = \sqrt{13 \sin^{2} \alpha - 9 \sin^{2} \alpha} = \sqrt{4 \sin^{2} \alpha} = 2 \sin \alpha$.Let $P(h, k)$ be the point. In $\Delta ABP$,where $B$ is the point of contact,$\angle APB = \alpha$ and $\angle ABP = 90^{\circ}$.Thus,$\sin \alpha = \frac{AB}{AP} = \frac{r}{AP} = \frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}$.$\Rightarrow \sqrt{(h+2)^{2}+(k-3)^{2}} = 2$.Squaring both sides,$(h+2)^{2}+(k-3)^{2} = 4$.$h^{2}+4h+4+k^{2}-6k+9 = 4$.$h^{2}+k^{2}+4h-6k+9 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+4x-6y+9=0$.

The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$ is $2 \alpha$. The equation of the locus of the point $P$ is

Similar Questions

Which of the following is a homosporous pteridophyte?

Cork cambium is a

Which substance is in higher concentration in blood than in glomerular filtrate?

The $pH$ of water is $7$. When a substance $Y$ is dissolved in water,the $pH$ becomes $13$. The substance $Y$ is a salt of

Vedclass Test Series

Exam Paper Generator

Online Exam Module