# Tangent to circle

The circle $x^2+y^2-4y+3=0$ passes through the points $(0,1),(-\frac{24}{25},\frac{43}{25}),(1,2)$. Its Centre is $(0,2)$ and its radius is $1$. I am asked to find the tangent to the circle through the point $(1,2$). This seems like a trivial problem but I have hit something which is not clear to me. I write the equation of the tangent as $y-2=m(x-1)$ which gives $mx-y+2-m=0$ as the tangent. I then calculate the perpindicular distance from the centre $(0,2)$ to the tangent and equate this distance to the radius $1$. This gives $$\frac{m(0)-1(2)+(2-m)}{\sqrt{(1+m^2)}}=1$$ which in turn gives $$-\frac{m}{\sqrt{(1+m^2)}}=1$$ which gives $$m^2=1+m^2$$

I suppose its correct if $m = \infty$ (which it is as the tangent is $x=1$) but this answer does not roll out of the equations. What have i done wrong here?

• Divide $m^2=1+m^2$ by $m^2$ to get $1/m=0$, thereby from the equation of the tangent, $(y-2).1/m=x-1$, $x=1$ is the required tangent. Does it answer your question..? – Tapu Mar 11 '13 at 11:54