What are the equations of the tangents drawn from the point (0,1) to the circle #x^2+y^2-2x+4y=0#?

1 Answer
Nov 4, 2017

The equations of tangents are #y=2x+1# and #x+2y-2=0#

Explanation:

A point #(x_1,y_1)# is outside a circle #x^2+y^2+2gx+2fy+c=0#, if #x_1^2+y_1^2+2gx_1+2fy_1+c>0#.

Here circle is #x^2+y^2-2x+4y=0# and #0^2+1^2-2xx0+4xx1=5>0#, hence #(0,1)# is outside the circle #x^2+y^2-2x+4y=0# and hence, we can have two tangents.

Equation of line with slope #m# and passing through #(0,1)# is #y-1=m(x-0)# or #y=mx+1#. Putting value of #y# from this in the equation of circle we get

#x^2+(mx+1)^2-2x+4(mx+1)=0#

or #x^2+m^2x^2+2mx+1-2x+4mx+4=0#

or #x^2(1+m^2)+x(6m-2)+5=0#

As we have a quadratic equation in #x#, the line #y=mx+1# would be tangent if it cuts the circle only at one point, which would be true if discriminant is zero or

#(6m-2)^2-20(1+m^2)=0#

or #36m^2-24m+4-20-20m^2=0#

or #16m^2-24m-16=0# or #2m^2-3m-2=0#

or #(2m+1)(m-2)=0#

i.e. #m=2# or #m=-1/2#

and equation of tangents are #y=2x+1# or #y=-1/2x+1# i.e. #x+2y-2=0#

graph{(x+2y-2)(2x-y+1)(x^2+y^2-2x+4y)(x^2+(y-1)^2-0.01)=0 [-4.69, 5.31, -3.5, 1.5]}