Question

Find the equation of the tangents to the circle $x^2+y^2-2x-4y-4=0$ which are (i) parallel,
(ii) perpendicular to the line $3x-4y-1=0$

Answer 1

(i)

Let the Equation of tangent parallel to $3x-4y-1=0$ be $3x-4y+k=0$

Now, Radius = Distance of center from the line

$\sqrt{1^2+2^2-(-4)}=\left|\frac{3\left(1\right)-4\left(2\right)+k}{\sqrt{9+16}}\right|\Rightarrow 3\ast 5=3-8+k\Rightarrow k=20$

Hence, Equation of tangent parallel to $3x-4y-1=0$
is $3x-4y+20=0$

(ii)

Let the Equation of tangent perpendicular to $3x-4y-1=0$ be $4x+3y+p=0$

Now, Radius = Distance of center from the line

$\sqrt{1^2+2^2-(-4)}=\left|\frac{4\left(1\right)+3\left(2\right)+p}{\sqrt{16+9}}\right|\Rightarrow 3\ast 5=4+6+p\Rightarrow p=5$

Hence, Equation of tangent perpendicular to $3x-4y-1=0$

is $4x+3y+5=0$