Question

If the line $y = mx + c$ is a common tangent to the hyperbola $\dfrac{{{x^2}}}{{100}} - \dfrac{{{y^2}}}{{64}} = 1$ and the circle ${x^2} + {y^2} = 36$, then which one of the following is true?
A. $4{c^2} = 369$
B. ${c^2} = 369$
C. $8m + 5 = 0$
D. $5m = 4$

Answer 1

Hint: Find out the conditions of the line $y = mx + c$ for being a tangent to the given hyperbola and the given circle. Substituting $y = mx + c$ in the given equation of the hyperbola, a quadratic equation will be found which must have two equal roots. Find out the discriminant and set an equation equating the discriminant to zero. A relation between $m$ and $c$ will be obtained. This is the condition of the given line for being a tangent to the given hyperbola. The line will be a tangent to the circle if the length of perpendicular from the center of the circle to the line is equal to the length of its radius. Using this, find another equation and simplify it to obtain the condition of the given line for being a tangent to the given circle. Solve the obtained two relationships between $m$ and $c$ to get the desired result.

Formula Used:
Discriminant of a quadratic equation $a{x^2} + bx + c = 0$ is given by $D = {b^2} - 4ac$
The length of perpendicular from origin to the line $ax + by + c = 0$ is $\dfrac{{\left| c \right|}}{{\sqrt {{a^2} + {b^2}} }}$ units.

Complete step by step solution:
The given equation of the hyperbola is $\dfrac{{{x^2}}}{{100}} - \dfrac{{{y^2}}}{{64}} = 1.....(i)$
and the given equation of the circle is ${x^2} + {y^2} = 36.....(ii)$
and the given equation of the line is $y = mx + c$, which can be written as $mx - y + c = 0....(iii)$
Find out the condition for the line $y = mx + c$ to be a tangent to the given hyperbola.
From the equations $(i)$ and $(iii)$, we get
      $\dfrac{{{x^2}}}{{100}} - \dfrac{{{{\left( {mx + c} \right)}^2}}}{{64}} = 1$
$\begin{array}{l} \Rightarrow \dfrac{{16{x^2} - 25{{\left( {mx + c} \right)}^2}}}{{1600}} = 1\\ \Rightarrow 16{x^2} - 25{\left( {mx + c} \right)^2} = 1600\\ \Rightarrow 16{x^2} - 25\left( {{m^2}{x^2} + 2mcx + {c^2}} \right) = 1600\\ \Rightarrow 16{x^2} - 25{m^2}{x^2} - 50mcx - 25{c^2} - 1600 = 0\\ \Rightarrow \left( {16 - 25{m^2}} \right){x^2} - 50mcx - 25\left( {{c^2} + 64} \right) = 0.....\left( {iv} \right)\end{array}$
If the equation $\left( {iv} \right)$ has equal roots, then the given line will intersect the given hyperbola at one point only and hence it will be the tangent to the hyperbola.
For equal roots, the discriminant of the quadratic equation $\left( {iv} \right)$ will be equal to zero.
Discriminant ${\left( { - 50mc} \right)^2} - 4\left( {16 - 25{m^2}} \right)\left\{ { - 25\left( {{c^2} + 64} \right)} \right\} = 0$
$\begin{array}{l} \Rightarrow 2500{m^2}{c^2} + 100\left( {16 - 25{m^2}} \right)\left( {{c^2} + 64} \right) = 0\\ \Rightarrow 100\left\{ {25{m^2}{c^2} + \left( {16 - 25{m^2}} \right)\left( {{c^2} + 64} \right)} \right\} = 0\\ \Rightarrow 25{m^2}{c^2} + \left( {16 - 25{m^2}} \right)\left( {{c^2} + 64} \right) = 0\\ \Rightarrow 25{m^2}{c^2} + 16{c^2} - 25{m^2}{c^2} + 1024 - 1600{m^2} = 0\\ \Rightarrow 16{c^2} + 1024 - 1600{m^2} = 0\\ \Rightarrow {c^2} + 64 - 100{m^2} = 0\\ \Rightarrow {c^2} = 100{m^2} - 64\\ \Rightarrow {c^2} = 4\left( {25{m^2} - 16} \right)\\ \Rightarrow c = \pm 2\sqrt {25{m^2} - 16} .......\left( v \right)\end{array}$
This is the condition.
Find out the condition for the line $y = mx + c$ to be a tangent to the given circle.
The line $(iii)$ will be a tangent to the circle $(ii)$ if the length of perpendicular from the center $\left( {0,0} \right)$ of the circle on the line is equal to the length of the radius of the circle.
The length of perpendicular from the center $\left( {0,0} \right)$ of the circle $(ii)$ on the line $(iii)$$ = \dfrac{{\left| c \right|}}{{\sqrt {{m^2} + 1} }}$ units
and the length of the radius of the circle is $6$ units.
So, $\dfrac{{\left| c \right|}}{{\sqrt {{m^2} + 1} }} = 6$
$\begin{array}{l} \Rightarrow \left| c \right| = 6\sqrt {{m^2} + 1} \\ \Rightarrow c = \pm 6\sqrt {{m^2} + 1} .......\left( {vi} \right)\end{array}$
This is the condition.
From equations $\left( v \right)$ and $\left( {vi} \right)$, we get
$ \pm 2\sqrt {25{m^2} - 16} = \pm 6\sqrt {{m^2} + 1} $
Cancel $ \pm 2$ from both sides, we get
$ \Rightarrow \sqrt {25{m^2} - 16} = 3\sqrt {{m^2} + 1} $
Square both sides of the equation.
$\begin{array}{l}\\ \Rightarrow 25{m^2} - 16 = 9\left( {{m^2} + 1} \right)\\ \Rightarrow 25{m^2} - 16 = 9{m^2} + 9\\ \Rightarrow 25{m^2} - 9{m^2} = 16 + 9\\ \Rightarrow 16{m^2} = 25\\ \Rightarrow {m^2} = \dfrac{{25}}{{16}}\\ \Rightarrow m = \pm \sqrt {\dfrac{{25}}{{16}}} \\ \Rightarrow m = \pm \dfrac{5}{4}\end{array}$
Putting the values of $m$ in the equation $\left( {vi} \right)$, we get
$c = \pm 6\sqrt {{{\left( { \pm \dfrac{5}{4}} \right)}^2} + 1} $
   $\begin{array}{l} = \pm 6\sqrt {\dfrac{{25}}{{16}} + 1} \\ = \pm 6\sqrt {\dfrac{{25 + 16}}{{16}}} \\ = \pm 6\sqrt {\dfrac{{41}}{{16}}} \end{array}$
$\begin{array}{l}\therefore {c^2} = 36 \times \dfrac{{41}}{{16}} = 9 \times \dfrac{{41}}{4} = \dfrac{{369}}{4}\\ \Rightarrow 4{c^2} = 369\end{array}$

Option ‘A’ is correct

Note: If there is a common tangent to two curves, then there is a point of intersection of the two curves, through which the tangent passes that’s why the quadratic equation (iv) must have two equal roots. If a quadratic equation has two equal roots, then its discriminant should be equal to zero. The distance between the center of a circle and its tangent must be equal to the length of its radius.