Equation of a line tangent to circumference Discover the general equation of the tangent line to the circumference $x^2 + y^2 - 2x + 4y + 1 = 0$ by the point $(3,4)$. NO CALCULUS.


*
  
*by the circumference equation i discovered that $C(1, -2)$ and $r=2$
  
*with the point $P(3,4)$ I put in the line equation:

$$(y - yo) = m (x - xo)$$
    $$y - 4 = mx - 3m$$ 
    $$mx - y + 4 - 3m = 0$$
    
*with the equation and the point of the circumference, I put them in the distance between point and line equation:
    
*


$$\frac{|a x + by + c|} { \sqrt{a² + b²}} = 2$$

$$\frac{|m + (-2)(-1) + 4 - 3m|}{ \sqrt{(m)² + (-1)²}} = 2$$
$$\frac{|-2m + 6| }{ \sqrt{m² + 1} }= 2$$
$$\left(\frac{|-2m + 6| }{\sqrt{m² + 1
}}\right)^2 = 2^2$$
$$4m² - 24m + 36 = 4m² + 4$$
$$m = \frac{3}{4}$$


    
*With this i found the equation: $\frac{3}{4}x - y = 0$


Wolfram graphic: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eohl0i7ciu3
Thanks everybody!
 A: Your equation describes a circle of radius two centered at the point $(1,-2)$. A line through a point $(x,y)$ on the circle is tangent to the circle if and only if it is perpendicular to the line from the center to the point $(x,y)$ and this line has slope $(y+2)/(x-1)$.  Thus, you could simultaneously solve the equations
$$\frac{y-4}{x-3}=-\frac{x-1}{y+2} \: \text { and } \: x^2-2 x+y^2+4 y+1=0.$$

A: 

*
  
*by the circumference equation i discovered that $C(1, -2)$ and $r=2$
  
*with the point $P(3,4)$ I put in the line equation:

$$(y - yo) = m (x - xo)$$
    $$y - 4 = mx - 3m$$ 
    $$mx - y + 4 - 3m = 0$$
    
*with the equation and the point of the circumference, I put them in the distance between point and line equation:
    
*


$$\frac{|a x + by + c|} { \sqrt{a² + b²}} = 2$$

$$\frac{|m + (-2)(-1) + 4 - 3m|}{ \sqrt{(m)² + (-1)²}} = 2$$
$$\frac{|-2m + 6| }{ \sqrt{m² + 1} }= 2$$
$$\left(\frac{|-2m + 6| }{\sqrt{m² + 1
}}\right)^2 = 2^2$$
$$4m² - 24m + 36 = 4m² + 4$$
$$m = \frac{3}{4}$$


    
*With this i found the equation: $\frac{3}{4}x - y = 0$


Wolfram graphic: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eohl0i7ciu3
A: Method 1 ( I am not sure whether you can use vector analysis, if not, see Method 2)
The equation of circle is $(x-1)^2+(y+2)^2=4$. $O=(1,-2), M=(3,4)$ Then points of the circle has form $P=(1+2cos(\theta),-2+2sin(\theta)),\theta \in [0,2\pi)$. Then the $OP=(2cos(\theta),2sin(\theta))$, $PM=(-2+2cos(\theta),-6+2sin(\theta))$. Since $OP$ is perpendicular to $PM$, we have $PM \cdot OP=0$, then we get $3sin(\theta )+cos(\theta )=1$. Then combined with $sin^2(\theta )+cos^2(\theta )=1$, we can get two points (the system can be solved easily $\sin(\theta_1)=0, \sin(\theta_2)=\frac{3}{5}$). Then we get two lines.
Method 2
We observe $P_1=(3,-2)$ is on the circle, hence one tangent line must be $x=3$. Then the other point $P_2=(x,y)$ must be symmetric w.r.t $P_1$ in terms of line passing through $OM$. 
Write down the line $y=x+1$, solve for $P_2$. Using $P_1P_2$ is perpendicular to OM, the middle point of $P_1P_2$ is on line $OM$. (you just need to solve to linear equation, instead of a quadratic equation), you get 
$$ \frac{y+2}{x-3}=-1 \quad \text{and} \quad \frac{x+3}{2}+1=\frac{y-2}{2}.$$ Solve for $(x,y)$, we get the other tangent line.
