# Geo4.31

Find the equations of the tangents from the point(0,1) to the circle $x^{2}+y^{2}-2x-6y+6=0\,$

Equation of the circle is

$x^{2}+y^{2}-2x-6y+6=0\,$

Its centre is (1,3) and radius is [/itex]\sqrt{1+9-6}=2\,[/itex]

Equation of any line through (0,1) is

$y-1=m(x-0)\,$

$mx-y+1=0\,$

If this touches the circle,the length of the pependicular from the centre(1,3) on the equation above(line) is numerically equal to 2.

Hence

$|{\frac {m(1)-3+1}{{\sqrt {m^{2}+1}}}}|=2\,$

$(m-2)^{2}=4(m^{2}+1)\,$

$m=0,{\frac {4}{3}}\,$

Substituting the value in the equation,the tangent equations are

y-1=0, 4x+3y-3=0