Question

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

Answer 1

Given circle is S = $x^2+y^2-2x+4y=0$.........................(i)
Let $P\equiv (0,1)$ For point P, $S_1=0^2+1^2-2.0+4.1=5$
Clearly P lies outside the circle and $T\equiv x.0+y.1-(x+0)+2(y+1)$ i.e. $T\equiv -x+3y+2$
Now equation of pair of tangents from P(0, 1) to circle (1) is $S{S}_1=T^2$
or $5(x^2+y^2-2x+4y)=(-x+3y+2{)}^2$
or $5x^2+5y^2-10x+20y)=x^2+9y^2+4-6xy-4x+12y$
or $4x^2-4y^2-6x+8y+6xy-4=0$
or $2x^2-2y^2+3xy-3x+4y-2=\mathrm{0...............}\left(ii\right)$
Separate equation of pair of tangents From (ii) $2x^2+3(y-1)x-2(2y^2-4y+2)=0$
$\therefore x=\frac{-3(y-1)\pm \sqrt{9(y-1{)}^2+8(2y^2-4y+2)}}{4}$
or $4x+3y-3=\pm \sqrt{25y^2-50y+25}=\pm 5(y-1)$
$\therefore$ Separate equations of tangents are $2x-y+1=0$ and $x+2y-2=0$