Question

If a line $y=mx+c$ is a tangent to the circle $(x-3{)}^2+y^2=1$ and it is perpendicular to a line $L_1,$ where $L_1$ is the tangent to the circle $x^2+y^2=1$ at the point $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),$ then

• A. $c^2+7c+6=0$
• B. $c^2-6c+7=0$
• C. $c^2+6c+7=0$
• D. $c^2-7c+6=0$

Answer 1

The correct option is C $c^2+6c+7=0$


Equation of $L_1:\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=1$
$\Rightarrow x+y=\sqrt{2}$
Now $L\perp L_1$
$\Rightarrow m\times m_1=-1$
$\Rightarrow m\times -1=-1$
$\Rightarrow m=1$
$\therefore$ Slope of $L=1$
$\Rightarrow$ Equation of $L:y=x+c$
Now as $L$ is tangent to $S:(x-3{)}^2+y^2=1,$
${\perp }^r$ distance from $O$ to $L=$ radius
$\Rightarrow \left|\frac{c+3}{\sqrt{2}}\right|=1$
$\Rightarrow |c+3|=\sqrt{2}$
$\Rightarrow c^2+9+6c=2$
$\Rightarrow c^2+6c+7=0$