Question

The circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$ and hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$intersect at the points $A$ and $B$. The equation of a common tangent with a positive slope to the circle as well as to the hyperbola is

• A. $2\mathrm{x}–\sqrt{5}\mathrm{y}–20=0$
• B. $2\mathrm{x}–\sqrt{5}\mathrm{y}+4=0$
• C. $3\mathrm{x}–4\mathrm{y}+8=0$
• D. $4\mathrm{x}–3\mathrm{y}+4=0$

Answer 1

The correct option is B

$2\mathrm{x}–\sqrt{5}\mathrm{y}+4=0$

Explanation for the correct option:

Step-1: Finding the equation of the tangent to the hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$.

The general equation of a hyperbola is $\frac{{\mathrm{x}}^2}{a^2}-\frac{{\mathrm{y}}^2}{b^2}=1$

Given equation of the hyperbola is $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$

Comparing the equations we get, $a^2=9,b^2=4$

We know that the equation of the tangent to the hyperbola $\frac{{\mathrm{x}}^2}{a^2}-\frac{{\mathrm{y}}^2}{b^2}=1$ is given by formula, $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}$, where $m$ is the slope of the tangent.

On substituting $a^2=9,b^2=4$ in the above formula, we get the equation of the given hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$as: $\mathrm{y}=\mathrm{mx}+\sqrt{9{\mathrm{m}}^2-4}$.

step-2: Finding the center of the given circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$

Given equation of the circle is: ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$.

Re-writing this equation, we get:

$$\begin{gathered} x^2+y^2-8x=0 \\ \Rightarrow x^2-2.4.x+4^2+y^2=4^2\left[\text{adding}{4}^2\text{on both sides}\right] \\ \Rightarrow {\left(x-4\right)}^2+{\left(y-0\right)}^2=4^2 \end{gathered}$$

We know that the center and radius of the circle given by the equation $\begin{array}{rcl}{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2& =& r^2\end{array}$ are: $\left(x_1,y_1\right)$ and $r$ respectively.

Therefore, for the given circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$ i.e. ${\left(x-4\right)}^2+{\left(y-0\right)}^2=0$, we get:$x_1=4,y_1=0$. So, the center of the circle is $\left(x_1,y_1\right)=(4,0)$.

Step-3: Finding the distance of the tangent from the center of the circle

Formula to be used: We know that the distance of the straight line $Ax+By+C=0$ from the point $\left(x_1,y_1\right)$ is given by: $D=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$.

Here, the equation of the tangent is $\mathrm{y}=\mathrm{mx}+\sqrt{9{\mathrm{m}}^2-4}$ i.e. $\mathrm{mx}-y+\sqrt{9{\mathrm{m}}^2-4}=0$. So, on comparing with $Ax+By+C=0$, we get: $A=m,B=-1,C=\sqrt{a^2{m}^2-b^2}$.

On substituting the given values $A=m,B=-1,C=\sqrt{9m^2-4}$ in $D=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$, we get the distance of the tangent to the given hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1$ from the center of the given circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$ as: $D=\left|\frac{4\mathrm{m}+\sqrt{9{\mathrm{m}}^2-4}}{\sqrt{1+{\mathrm{m}}^2}}\right|$.

Step-3: Finding the value of $m$

Now since, the distance of the tangent to the circle from its center equals the radius of the circle and we are taking the tangent that is common to both the hyperbola and the circle and the radius of the given circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$ is $r=4$, we get:

$$\begin{gathered} D=4 \\ \Rightarrow \left|\frac{4\mathrm{m}+\sqrt{9{\mathrm{m}}^2-4}}{\sqrt{1+{\mathrm{m}}^2}}\right|=4 \\ \Rightarrow 49{\mathrm{m}}^4+104{\mathrm{m}}^2-400=0 \\ \Rightarrow {\mathrm{m}}^2=\frac{4}{5} \\ \Rightarrow \mathrm{m}=\pm \frac{2}{\sqrt{5}} \end{gathered}$$

Step 4: Finding the equation of tangent after substituting the value of $m$

Hence, the equation of common tangent is

$\begin{array}{rcl}\mathrm{y}& =& \mathrm{mx}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2-{\mathrm{b}}^2}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{9{\mathrm{m}}^2-4}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{9{\left(\frac{2}{\sqrt{5}}\right)}^2-4}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{9\left(\frac{4}{5}\right)-4}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{\frac{36}{5}-4}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{\frac{36-20}{5}}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{\frac{16}{5}}\\ & =& \frac{2}{\sqrt{5}}\mathrm{x}+\frac{4}{\sqrt{5}}\\ & \Rightarrow & \sqrt{5}\mathrm{y}=2\mathrm{x}+4\end{array}$

Therefore, the equation of tangent is $\begin{array}{rcl}2\mathrm{x}-\sqrt{5}\mathrm{y}+4& =& 0\end{array}$

Hence, option (B) is the correct answer