Question

If $y=3x$ is the tangent to a circle with centre $\left(1,1\right)$, then the other tangent drawn through $\left(0,0\right)$ to the circle is

• A. $3y=x$
• B. $y=-3x$
• C. $y=2x$
• D. $y=-2x$

Answer 1

The correct option is A

$3y=x$

Explanation for the correct answer.

Step 1. Find the radius of the circle.

The perpendicular distance between a point $\left(x_1,y_1\right)$ and a line $ax+by+c=0$ is given as: $d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$.

The equation $y=3x$ or $-3x+y=0$ is the tangent to the circle with center $\left(1,1\right)$.

Now the perpendicular distance between the center of the circle and the tangent is the radius of the circle.

So the distance between the point $\left(1,1\right)$ and the line $-3x+y=0$ is:

$\begin{array}{rcl}r& =& \frac{\left|-3\times 1+1\right|}{\sqrt{{\left(-3\right)}^2+1^2}}\\ & =& \frac{\left|-3+1\right|}{\sqrt{9+1}}\\ & =& \frac{\left|-2\right|}{\sqrt{10}}\\ & =& \frac{2}{\sqrt{10}}\end{array}$

So the radius of the circle is $r=\frac{2}{\sqrt{10}}$

Step 2. Find the equation of the other tangent.

Let the equation of the other tangent passing through $\left(0,0\right)$ be $y=mx$ and in standard form be $-mx+y=0$.

Now the perpendicular distance between the center and the tangent line is the radius of the circle.

So the distance between the point $\left(1,1\right)$ and the line is $-mx+y=0$ is $r=\frac{2}{\sqrt{10}}$.

$$\begin{gathered} \frac{\left|-m\times 1+1\right|}{\sqrt{{\left(-m\right)}^2+1^2}}=\frac{2}{\sqrt{10}} \\ \Rightarrow \frac{\left|-m+1\right|}{\sqrt{m^2+1}}=\frac{2}{\sqrt{10}} \end{gathered}$$

Now square both sides and solve for $m$.

$$\begin{gathered} {\left(\frac{\left|-m+1\right|}{\sqrt{m^2+1}}\right)}^2={\left(\frac{2}{\sqrt{10}}\right)}^2 \\ \Rightarrow \frac{{(-m+1)}^2}{{\left(\sqrt{m^2+1}\right)}^2}=\frac{2^2}{{\left(\sqrt{10}\right)}^2} \\ \Rightarrow 10\left(m^2-2m+1\right)=4\left(m^2+1\right) \\ \Rightarrow 10m^2-20m+10=4m^2+4 \\ \Rightarrow 10m^2-20m+10-4m^2-4=0 \\ \Rightarrow 6m^2-20m+6=0 \\ \Rightarrow 2\left(3m^2-10m+3\right)=0 \\ \Rightarrow 3m^2-10m+3=0 \end{gathered}$$

Now using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ the quadratic equation can be solved for $m$ as:

$\begin{array}{rcl}m& =& \frac{-\left(-10\right)\pm \sqrt{{(-10)}^2-4\times 3\times 3}}{2\times 3}\\ & =& \frac{10\pm \sqrt{100-36}}{6}\\ & =& \frac{10\pm \sqrt{64}}{6}\\ & =& \frac{10\pm 8}{6}\end{array}$

So either

$\begin{array}{rcl}m& =& \frac{10+8}{6}\\ & =& \frac{18}{6}\\ & =& 3\end{array}$

or

$\begin{array}{rcl}m& =& \frac{10-8}{6}\\ & =& \frac{2}{6}\\ & =& \frac{1}{3}\end{array}$

So when $m=3$ the equation of the tangent is $y=3x$.

And when $m=\frac{1}{3}$ the equation of the tangent is:

$$\begin{gathered} y=\frac{1}{3}x \\ \Rightarrow 3y=x \end{gathered}$$

So the equation of the other tangent is $3y=x$.

Hence, the correct option is A.