Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the combined equation of two tangents drawn from $P(x_1,y_1)$ to the circle $x^2+y^2 = a^2$. Point $P$ lies outside the circle.

share|improve this question
3  
Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Find..."), not a request for help, so please consider rewriting it. –  Zev Chonoles Feb 7 '13 at 14:32
    
thanks!!! i will take care abt that now. –  user60576 Feb 7 '13 at 14:40

2 Answers 2

Sorry for the lack of a picture. You can show that the angle between the line from the origin to $(x_1,y_1)$ and a radial line at a point of tangency is

$$\tan{\theta} = \frac{\sqrt{x_1^2+y_1^2-a^2}}{a}$$

where $a$ is the radius of the circle centered at the origin. Let $\theta_1$ be the angle between the line from the origin to $(x_1,y_1)$ and the positive $x$ axis. Then the tangent points on the circle are given by

$$(a \cos{(\theta_1 \pm \theta)},a \sin{(\theta_1 \pm \theta)})$$

The combined equation of the tangent lines is then

$$y-y_1 = m_{\pm} (x-x_1)$$

where

$$m_{\pm} = \frac{y1-a \sin{(\theta_1 \pm \theta)}}{x_1-a \cos{(\theta_1 \pm \theta)}}$$

share|improve this answer
    
You might want to precise what $r$ is before your formula. I assume you are talking of the point $(r\cos\theta,r\sin\theta)$ on the circle, so maybe $r=a$? –  1015 Feb 7 '13 at 14:39
    
@julien: thanks for pointing out the typo. Yeah, I had $r$ in mind when I meant $a$. –  Ron Gordon Feb 7 '13 at 14:41

Let $(h,k)$ be any point on the tangent.

So,the equation of the line passing through $(x_1,y_1), (h,k)$ is $$\frac{y-y_1}{x-x_1}=\frac{k-y_1}{h-x_1}$$

$$\text { or, } x(k-y_1)+y(x_1-h)+hy_1-kx_1=0$$

Now, as the line is tangent to the circle, the perpendicular distance of the line form the centre $(0,0)$ of circle is the length of the radius $=a$ assuming $a>0$

So, $$\frac{|hy_1-kx_1|}{\sqrt{(k-y_1)^2+(x_1-h)^2}}=a$$

Squaring we get, $$(hy_1-kx_1)^2=a^2\{(k-y_1)^2+(x_1-h)^2\}$$

Hence, the locus of $(h,k)$ is $$(xy_1-yx_1)^2=a^2\{(y-y_1)^2+(x_1-x)^2\}$$ which is clearly the equations of the pair of tangents of $x^2+y^2=a^2$ from $(x_1,y_1)$


Alternatively, from the equation of the line $x(y_1-k)=y(x_1-h)+hy_1-kx_1$

Putting the value of $x,$ in the equation of the circle,

$$\{y(x_1-h)+hy_1-kx_1\}^2+y^2(y_1-k)^2=a^2(y_1-k)^2$$

$$\implies y^2\{(x_1-h)^2+(y_1-k)^2\}+2y(x_1-h)(hy_1-kx_1)+(hy_1-kx_1)^2-a^2(y_1-k)^2=0$$

This is a quadratic equation in $y$

Now, as the line is tangent to the circle, both the roots should be same i.e., the discriminant must be $0$

$$\{2(x_1-h)(hy_1-kx_1)\}^2-4\cdot\{(x_1-h)^2+(y_1-k)^2\}\cdot\{(hy_1-kx_1)^2-a^2(y_1-k)^2\}=0$$

$$\implies(y_1-k)^2a^2\{(k-y_1)^2+(h-x_1)^2\}=(y_1-k)^2(hy_1-kx_1)^2$$

As this is true for all values of $k$ even when $k\ne y_1$

So, $$a^2\{(k-y_1)^2+(h-x_1)^2\}=(hy_1-kx_1)^2$$

Hence, the locus of $(h,k)$ is $$(xy_1-yx_1)^2=a^2\{(y-y_1)^2+(x_1-x)^2\}$$ which is clearly the equations of the pair of tangents of $x^2+y^2=a^2$ from $(x_1,y_1)$

Observe from the Article#$160$ here, there will exactly two, one or zero real tangents from $(x_1,y_1)$ to the given circle according as $(x_1,y_1)$ lies outside, on or inside/within the circle.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.