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. $c^2-7c+6=0$
• D. $c^2+6c+7=0$

Answer 1

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

The equation of tangent $\left(L_1\right)$ to $x^2+y^2=1$ at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ is
$x{x}_1+y{y}_1-1=0\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}-1=0\Rightarrow x+y-\sqrt{2}=0$
Slope of $L_1$ is
$m_{L_1}=-1$
Slope of perpendicular line is
$m=1$

Given line is $y=mx+c$, so
$y=x+c$ is tangent to $(x-3{)}^2+y^2=1$
Centre and radius are
$C=(3,0),r=1$
We know that the perpendicular distance of tangent from centre is equal to radius, so
$\left|\frac{3+c}{\sqrt{2}}\right|=1\Rightarrow c+3=\pm \sqrt{2}$
Squaring both sides, we get
$\Rightarrow c^2+6c+9=2\therefore c^2+6c+7=0$