# Find the angle between two tangents from an external point $(x_1,y_1)$ to the circle $x^2+y^2=a^2$ .

Find the angle between two tangents from an external point $(x_1,y_1)$ to the circle $x^2+y^2=a^2$ .

Below is my attempt at the problem:

$x^2+y^2=a^2$,Differentiating we get $\dfrac{dy}{dx}=\dfrac{-x}{y}.$

Hence slope will be $\dfrac{-a_1}{b_1}$ at the point $(a_1,b_1)$ for the first tangent.

Similarly slope will be $\dfrac{-a_2}{b_2}$ at the point $(a_2,b_2)$ for the second tangent.

Now $a_1^2+b_1^2=a^2,a_2^2+b_2^2=a^2\implies (a_2-a_1)(a_2+a_1)+(b_2-b_1)(b_2+b_1)=0$

$\dfrac{-a_1}{b_1}=\dfrac{y_1-b_1}{x_1-a_1}\implies a_1x_1+y_1b_1=a_1^2+b_1^2$

Similarly $a_2x_2+y_2b_2=a_2^2+b_2^2$

Now to find the angle between them we have to evaluate $\tan\alpha=\dfrac{a_1b_2-a_2b_1}{b_1b_2+a_1a_2}$

How to eliminate $a_1,b_1,a_2,b_2$? from here? Please help.

• But this angle is not unique...it's okay to depend on the contact points Dec 29 '17 at 14:15

For $a>0$ it's just $$2\arcsin\frac{a}{\sqrt{x_1^2+y_1^2}}.$$

Indeed, let $A(x_1,y_1)$, $O(0,0)$ and $B$ be one of touching points.

Thus, the needed angle it's $2\measuredangle OAB$, $OA=\sqrt{x_1^2+y_1^2}$, $OB=a$ and $\measuredangle ABO=90^{\circ}$.

• How did you get that? Will you please explain Dec 29 '17 at 14:17
• $$O(0,0);P(x_1,y_1);|OP|=?$$ Dec 29 '17 at 14:18
• @New_User I added something. See now. Dec 29 '17 at 14:26
• Yes it is clear now for me Thank you Dec 29 '17 at 14:28
• You are welcome! Dec 29 '17 at 14:30

the equation any tangent to $x^2+y^2=a^2$ is $$y=mx\pm a\sqrt{m^2+1}$$

$$\iff(y-mx)^2=a^2(m^2+1)\iff m^2(a^2-x^2)+2mxy+a^2-y^2=0$$

Now this pair of tangents pass through $(x_1,y_1)$

$$m^2(a^2-x_1^2)+2mx_1y_1+a^2-y_1^2=0$$ whose roots are $m_1,m_2$

$$\implies m_1+m_2=\dfrac{-2x_1y_1}{a^2-x_1^2}$$ $$m_1m_2=\dfrac{a^2-y_1^2}{a^2-x_1^2}$$

We need to find $$\dfrac{|m_2-m_1|}{1+m_1m_2}=\dfrac{\sqrt{(m_2+m_1)^2-4m_1m_2}}{1+m_1m_2}$$