Geo4.31

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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 </math>\sqrt{1+9-6}=2\,</math>

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


Main Page:Geometry:Circles