Question

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

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

Answer 1

The correct option is B $2x-\sqrt{5}y+4$

Given:Circle:C:$x^2+y^2-8x=0$ is of the form $x^2+y^2+2gx+2fy+c=0$

where $2gx=-8x\Rightarrow g=-4$ and $f=0,c=0$

Radius of the cirlce $=\sqrt{g^2+f^2-c}=\sqrt{{(-4)}^2+0+0}=4$ units

and center of the circle is $(4,0)$

Hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$

The circle and hyperbola intersect at points $A$ and $B$

Now, the equation of the tangent to the hyperbola is

$\frac{x\sec \theta }{3}+\frac{y\tan \theta }{2}-1=0$

Perpendicular distance from the center $(4,0)$ of the circle to the tangent to the hyperbola is

$\frac{\frac{4\sec \theta }{3}+\frac{0\times \tan \theta }{2}-1}{\sqrt{\frac{{\sec }^2\theta }{9}+\frac{{\tan }^2\theta }{4}}}=$radius of the circle

$\Rightarrow \frac{\frac{4\sec \theta }{3}-1}{\sqrt{\frac{4{\sec }^2\theta +9{\tan }^2\theta }{36}}}=4$

$\Rightarrow \frac{4\sec \theta -3}{3}\times \frac{6}{\sqrt{4{\sec }^2\theta +9{\tan }^2\theta }}=4$

$\Rightarrow \frac{(4\sec \theta -3)}{\sqrt{4{\sec }^2\theta +9{\tan }^2\theta }}=2$

Squaring both sides,we get

$\Rightarrow 16{\sec }^2\theta +9-24\sec \theta =4(4{\sec }^2\theta +9{\tan }^2\theta )$

$\Rightarrow 16{\sec }^2\theta +9-24\sec \theta =16{\sec }^2\theta +36{\tan }^2\theta$

$\Rightarrow 9-24\sec \theta =36{\tan }^2\theta$

$\Rightarrow 9-24\sec \theta =36({\sec }^2\theta -1)$

$\Rightarrow 9-24\sec \theta =36{\sec }^2\theta -36$

$\Rightarrow 36{\sec }^2\theta +24\sec \theta -36-9=0$

$\Rightarrow 36{\sec }^2\theta +24\sec \theta -45=0$

$\Rightarrow 12{\sec }^2\theta +8\sec \theta -15=0$ dividing by $3$

$\Rightarrow \sec \theta =\frac{-8\pm \sqrt{64+720}}{2\times 12}$

$=\frac{-8\pm \sqrt{64-4\times 12\times -15}}{2\times 12}$

$=\frac{-8\pm 28}{2\times 12}$

$\sec \theta =\frac{-8+28}{2\times 12},\frac{-8-28}{2\times 12}$

$\sec \theta =\frac{20}{2\times 12},\frac{-36}{2\times 12}$

$\sec \theta =\frac{5}{6},\frac{-3}{2}$

But $\tan \theta =\sqrt{{\sec }^2\theta -1}=\sqrt{{\left(\frac{5}{6}\right)}^2\theta -1}=\sqrt{\frac{25-36}{36}}$ is imaginary.

$\therefore \sec \theta =\frac{-3}{2}$

$\Rightarrow \tan \theta =\sqrt{{\sec }^2\theta -1}=\sqrt{{\left(\frac{-3}{2}\right)}^2\theta -1}=\sqrt{\frac{9-4}{4}}=\frac{\sqrt{5}}{2}$

substituting $\sec \theta =\frac{-3}{2}$ and $\tan \theta =\frac{\sqrt{5}}{2}$ in $\frac{x\sec \theta }{3}+\frac{y\tan \theta }{2}-1=0$ we get

$\frac{\frac{-3x}{2}}{3}+\frac{\frac{\sqrt{5}y}{2}}{2}-1=0$

or $\frac{-3x}{6}+\frac{\sqrt{5}y}{4}-1=0$

or $\frac{x}{2}-\frac{\sqrt{5}y}{4}+1=0$

or $2x-\sqrt{5}y+4$ is the required equation of the common tangent with positive slope to the circle as well as the hyperbola