Question

The equation of the tangents drawn from the point $(0,1)$ to the circle $x^2+y^2-2x+4y=0$ are:

Answer 1

A point $(x_1,y_1)$ is outside a circle $x^2+y^2+2gx+2fy+c=0$, if

$x_1^2+y_1^2+2g{x}_1+2f{y}_1+c>0$

Here circle is $x^2+y^2-2x+4y=0$ and $0^2+1^2-2\times 0+4\times 1=5>0$

Hence $(0,1)$ is outside the circle and we have two tangents

Equation of line with slope 'm' and passing through (0,1)

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

$y=mx+1$

Putting value of y from this in the equation of circle, we get

$x^2+(mx+1{)}^2-2x+4(mx+1)=0$

$x^2+m^2{x}^2+2mx+1-2x+4mx+4=0$

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

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

$(6m-2{)}^2-20(1+m^2)=0$

$36m^2-24m+4-20-20m^2=0$

$16m^2-24m-16=0$

$2m^2-3m-2=0$

$(2m+1)(m-2)=0$

the equation of tangents are

$2x-y+1=0,x+2y-2=0$