Question

The circle ${x^2} + {y^2} - 8x = 0$ and hyperbola $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$ intersect at points A and B.
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is,
$\left( a \right)2x - \sqrt 5 y - 20 = 0$
$\left( b \right)2x - \sqrt 5 y + 4 = 0$
$\left( c \right)3x - 4y + 8 = 0$
$\left( d \right)4x - 3y + 4 = 0$

Answer 1

Hint: In this particular question use the concept that the equation of tangent to the hyperbola is given as $y = mx + \sqrt {{{\left( {am} \right)}^2} - {b^2}} ,m > 0$ where and b are the length of semi major and semi minor axis, as the tangent is common to both the circle and the hyperbola so this tangent is also the tangent to the circle, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equations
$\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$................ (1) and ${x^2} + {y^2} - 8x = 0$................. (2)
Now the standard equation of hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ now compare with equation (1) we have,
$ \Rightarrow {a^2} = 9,{b^2} = 4$
Now as we know that the equation of tangent to the hyperbola is given as $y = mx \pm \sqrt {{{\left( {am} \right)}^2} - {b^2}} $ where and b are the length of semi major and semi minor axis.
Now for, m > 0 the equation of tangent becomes, $y = mx + \sqrt {{{\left( {am} \right)}^2} - {b^2}} $
So the equation of tangent becomes,
$ \Rightarrow y = mx + \sqrt {9{m^2} - 4} $
$ \Rightarrow mx - y + \sqrt {9{m^2} - 4} = 0$.................. (3)
Now as we know that the standard equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, where (-g, -f) and r = $\sqrt {{g^2} + {f^2} - c} $ is the center and the radius of the circle respectively.
So on comparing the above equation with equation 2 we have,
$ \Rightarrow 2g = - 8,2f = 0,c = 0$
$ \Rightarrow g = - 4,f = 0,r = \sqrt {16 + 0 - 0} = 4$
So the center of the circle is $\left( { - g, - f} \right) = \left( {4,0} \right)$ and radius of the circle is r = 4.
Now the tangent of the hyperbola is common to the circle, so the tangent also is the tangent to the circle.
Now as we know that the tangent of the circle always makes a 90 degree with the center of the circle.
Now as we know that the perpendicular drawn from the point $\left( {{x_1},{y_1}} \right)$ on the line ax + by + c = 0 is given as,
$d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
So the radius (r = 4) of the circle = perpendicular drawn from the center (4, 0) of the circle on the tangent of the circle, $mx - y + \sqrt {9{m^2} - 4} = 0$
$ \Rightarrow 4 = \dfrac{{\left| {4m + \sqrt {9{m^2} - 4} } \right|}}{{\sqrt {{m^2} + 1} }}$
Now simplify this we have,
$ \Rightarrow 4\sqrt {{m^2} + 1} = 4m + \sqrt {9{m^2} - 4} $
Now take square root on both sides and use the property that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,
$ \Rightarrow {\left( {4\sqrt {{m^2} + 1} } \right)^2} = {\left( {4m + \sqrt {9{m^2} - 4} } \right)^2}$
$ \Rightarrow 16{m^2} + 16 = 16{m^2} + 9{m^2} - 4 + 8m\sqrt {9{m^2} - 4} $
$ \Rightarrow 20 - 9{m^2} = 8m\sqrt {9{m^2} - 4} $
Now again take square roots on both sides we have,
$ \Rightarrow {\left( {20 - 9{m^2}} \right)^2} = {\left( {8m\sqrt {9{m^2} - 4} } \right)^2}$
$ \Rightarrow 400 + 81{m^4} - 360{m^2} = 576{m^4} - 256{m^2}$
$ \Rightarrow 495{m^4} + 104{m^2} - 400 = 0$
Let, ${m^2} = p$.......... (4)
$ \Rightarrow 495{p^2} + 104p - 400 = 0$
Now apply quadratic formula we have,
$ \Rightarrow p = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 495, b = 104, and c = -400
$ \Rightarrow p = \dfrac{{ - 104 \pm \sqrt {{{\left( {104} \right)}^2} - 4\left( {495} \right)\left( { - 400} \right)} }}{{2\left( {495} \right)}} = \dfrac{{ - 104 \pm \sqrt {802816} }}{{990}} = \dfrac{{ - 104 \pm 896}}{{990}}$
$ \Rightarrow p = \dfrac{{ - 104 \pm 896}}{{990}} = - \dfrac{{1000}}{{990}},\dfrac{{792}}{{990}}$
Now from equation (4) we have,
$ \Rightarrow {m^2} = - \dfrac{{1000}}{{990}},\dfrac{{792}}{{990}}$
So, ${m^2} = \dfrac{{ - 1000}}{{990}}$ cannot possible otherwise roots will be complex
So, ${m^2} = \dfrac{{792}}{{990}} = \dfrac{4}{5}$ is a possible case.
$ \Rightarrow m = \pm \sqrt {\dfrac{4}{5}} = \pm \dfrac{2}{{\sqrt 5 }}$
Now it is given in the question that slope is positive so,
$ \Rightarrow m = \dfrac{2}{{\sqrt 5 }}$
Now substitute this value in equation (3) we have,
$ \Rightarrow \dfrac{2}{{\sqrt 5 }}x - y + \sqrt {9{{\left( {\dfrac{2}{{\sqrt 5 }}} \right)}^2} - 4} = 0$
$ \Rightarrow \dfrac{2}{{\sqrt 5 }}x - y + \sqrt {9\left( {\dfrac{4}{5}} \right) - 4} = 0$
$ \Rightarrow \dfrac{2}{{\sqrt 5 }}x - y + \sqrt {\dfrac{{16}}{5}} = 0$
$ \Rightarrow \dfrac{2}{{\sqrt 5 }}x - y + \dfrac{4}{{\sqrt 5 }} = 0$
$ \Rightarrow 2x - \sqrt 5 y + 4 = 0$
So this is the required equation of the tangent.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the perpendicular drawn from the point $\left( {{x_1},{y_1}} \right)$ on the line ax + by + c = 0 is given as, $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ so simplify take the perpendicular from the center of the circle on the equation of tangent and equate it to the radius of the circle as above and simplify we will get the value of slope as we want the positive slope so neglect negative sign and substitute this value in the equation of tangent we will get the required common tangent.