# To find tangents to given circle from a point outside it

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.

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thanks!!! i will take care abt that now. –  user60576 Feb 7 '13 at 14:40

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

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

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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.

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