Question

Find equation of tangent to $x^2+y^2-2x+4y=0$ at (3, -1). Also find equation of tangent parallel to it.

Answer 1

We know the equation of tangent for a curve x${}^2$ +y${}^2$+2gx+2by+c=0

is S=0 where S is

$S$ :$\left(x{x}_1\right)+\left(y{y}_1\right)+g(x+x_1)+l(y+y_1)+c=0$

where ($(x_1,y_1)$ are points on curve at which tangent is drawn.

$\Rightarrow$ Equation of tangent at (3,-1) :

$\Rightarrow$$3x-y-1(2+3)+2(y-1)=0$

$\Rightarrow$ $3x-y-x-3+2y-2=0$

$\Rightarrow$ $3x+y-x-5=0$

$\Rightarrow$ $2x+y=5$.

Parallel tangent will obviously have some slope =$-2$,

$\therefore$ Its equation $y=-2x+c^1.$

radius of circle is $=\sqrt{g^2+b^2-c}$

$=\sqrt{1+4-0}$

$=\sqrt{5}.$

Centre is $(1,-2).$

and condition of tangency is :

$C^2=a^2(m^2+1)$

Here $C=C^1,a=radius=\sqrt{5}$

$m=-2$

$\Rightarrow$ $C^2=5(4+1)$

$\Rightarrow$$C^1=-5(\because +5isofothertangent.)$

$\therefore$ Eq. of parallel tangent : $y=-2x-5.$