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