Question

Show that two tangent can be drawn from the point$(9,0)$ to the circle $x^2+y^2=16$ ; also find the equation of the pair of tangents and the angle between them.

• A. $si{n}^{-1}\frac{4}{9}$
• B. $si{n}^{-1}\frac{\sqrt{65}}{9}$
• C. $(\sqrt{65}y+4x-36)$$\left(\sqrt{(}65\right)y-4x+36)$=0
• D. $(\sqrt{65}y+4x+36)$$(\sqrt{65}y-4x-36)$=0

Answer 1

Answer: (A), (C)

The correct options are
A $si{n}^{-1}\frac{4}{9}$
C $(\sqrt{65}y+4x-36)$$\left(\sqrt{(}65\right)y-4x+36)$=0

Equation of the circle $x^2+y^2=16$

Center, $C=(0,0)$ and Radius, $r=4$

Distance between $(9,0)$ and $(0,0)$,

$d=\sqrt{(0-9{)}^2+(0-0{)}^2}=(\sqrt{9}{)}^2=9$

$d=9>r$

since the point lies outside the circle.

$\therefore$ Two tangents can be drawn from $(9,0)$ to the circle.

Equation of tangent to the circle $x^2+y^2=0$ is,

$y=mx\pm a\left(\sqrt{1+m^2}\right)$

$y-mx=\pm a\left(\sqrt{1+m^2}\right)$

Squaring both sides,

$(y-mx{)}^2=(\pm a\left(\sqrt{1+m^2}\right){)}^2$

$(y-mx{)}^2=a^2(\sqrt{1+m^2}{)}^2$

$a=4$, substituting

$(y-mx{)}^2=4^2(1+m^2)$

$\therefore (y-mx{)}^2=16(1+m^2)$

This line passes through $(9,0)$,

Put, $x=9$ and $y=0$

$(-9m{)}^2=16+16m^2$

$81m^2-16m^2=16$

$65m^2=16$

$m=\sqrt{\frac{16}{65}}=\pm \frac{4}{\sqrt{65}}$

$m_1=+\frac{4}{\sqrt{65}}$ , $m_2=-\frac{4}{\sqrt{65}}$

$m_1+m_2=0$