Question

Equation of tangent drawn from origin to the circle $x^2+y^2-2rx-2hy+h^2=0$ are:

• A. $(h^2-r^2)x-2rhy=0$
• B. $y=0$
• C. $(h^2-r^2)x+2rhy=0$
• D. $noneofthese$

Answer 1

Answer: (C)

$(h^2-r^2)x+2rhy=0$

Pair of tangents from origin on the given circle is

$T^2={SS}_1\Rightarrow (0+0-r(x+0)-h(y+0)+h^2{)}^2=0(x^2+y^2-2rx-2hy+h^2\left)\right(h^2)\Rightarrow (-rx-hy+h^2)=x^2{h}^2+y^2{h}^2-2r{h}^{2}x-2h^{3}y+h^4\Rightarrow r^2{x}^2+h^2{y}^2+h^4+2(rhxy-h^{3}y-r{h}^{2}x)=x^2{h}^2+y^2{h}^2-2r{h}^{2}x-2h^{3}y+h^4\Rightarrow r^2{x}^2+2rhxy=x^2{h}^2\Rightarrow x^2(h^2-r^2)=2rhxy.\Rightarrow x\{x(h^2-r^2)-2rhy\}=0\therefore x=0andx(h^2-r^2)=2rhxyareequationsoftangent.$