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 $\left(\frac{1}{\surd 2},\frac{1}{\surd 2}\right)$; 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

$c^2+6c+7=0$

Explanation for the correct option:

Step 1. Find the slope of tangent to the circle:

Slope of tangent to circle $x^2+y^2=1$ at $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ is

$\frac{dy}{dx}=\frac{-x}{y}$

$\Rightarrow$$m_1=-1$

Given that $y=mx+c$ is a tangent to the circle ${(x-3)}^2+y^2=1$

$\because$Tangent is perpendicular to line $l_1$

$\therefore m_1.m=-1$

$\Rightarrow$ $m=-\frac{1}{m_1}$

$\Rightarrow$ $m=1$

Step 2. Apply tangency condition:

$\Rightarrow$ $\frac{\left|3+c\right|}{\sqrt{2}}=1$

$\Rightarrow$ $c=-3\pm \sqrt{2}$

$\Rightarrow$ $c^2+6c+9=2$

$\mathbf{\therefore }{\mathit{c}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{6}\mathit{c}\mathbf{}\mathbf{+}\mathbf{}\mathbf{7}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘C’ is Correct.