Question

If $a > 2b > 0$, then the positive value of m for which $y = mx - b\sqrt {1 + {m^2}}$ is a common tangent to ${x^2} + {y^2} = {b^2}$ and ${\left( {x - a} \right)^2} + {y^2} = {b^2}$ is:
A. $\dfrac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}$
B. $\dfrac{{\sqrt {{a^2} - 4{b^2}} }}{{2b}}$
C. $\dfrac{{2b}}{{a - 2b}}$
D. $\dfrac{b}{{a - 2b}}$

Answer 1

Hint: The given problem tests us on the concepts of coordinate geometry as well as conic sections. The problem requires us to the slope of the common tangent of two circles whose equation is given to us in the problem itself. We have to analyze the problem by drawing a figure for the situation and then plan our sequence of actions. We find the centers and radii of both the circles. Then, we equate the distance of the tangent from the circle with the radius of that circle as the radius and the tangent form a right angle.

Complete step by step answer:

The equations of circles given to us are: ${x^2} + {y^2} = {b^2}$ and ${\left( {x - a} \right)^2} + {y^2} = {b^2}$.

Now, we know that the standard equation of a circle whose centre is known to us as $\left( {\alpha ,\beta } \right)$ and the radius is r units is ${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}$. So, the centre of the circle ${x^2} + {y^2} = {b^2}$ is $\left( {0,0} \right)$ and radius is $b$ units. Also, the center of the circle ${\left( {x - a} \right)^2} + {y^2} = {b^2}$ is $\left( {a,0} \right)$ and the radius is $b$ units.

Now, we find the distance of the tangent from the centers of both the circles and equate these distances with the radii of the respective circles.

Now, we know that the distance of a line $ax + by + c = 0$ from a point $\left( {p,q} \right)$ is $\left| {\dfrac{{ap + bq + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.

So, the distance of the tangent $y = mx - b\sqrt {1 + {m^2}}$from center of the circle ${x^2} + {y^2} = {b^2}$, $\left( {0,0} \right)$ is $\left| {\dfrac{{\left( 0 \right) - m\left( 0 \right) - b\sqrt {1 + {m^2}} }}{{\sqrt {1 + {m^2}} }}} \right| = \left| { - b} \right|$.

Now, we know that a and b are positive constants. So, the distance of the tangent $y = mx - b\sqrt {1 + {m^2}}$from center of the circle ${x^2} + {y^2} = {b^2}$, $\left( {0,0} \right)$ is b units. Also, the radius of the circle ${x^2} + {y^2} = {b^2}$ is b units. Hence, this is true for every value of m.

Now, the distance of the tangent $y = mx - b\sqrt {1 + {m^2}}$from center of the circle ${\left( {x - a} \right)^2} + {y^2} = {b^2}$, $\left( {a,0} \right)$ is $\left| {\dfrac{{\left( 0 \right) - m\left( a \right) - b\sqrt {1 + {m^2}} }}{{\sqrt {1 + {m^2}} }}} \right| = \left| {\dfrac{{ - ma - b\sqrt {1 + {m^2}} }}{{\sqrt {1 + {m^2}} }}} \right|$.

Now, we know that the radius of the circle ${\left( {x - a} \right)^2} + {y^2} = {b^2}$ is b units.So, equating the distance of tangent from center of the circle with the radius of the circle, we get,

$\left| {\dfrac{{ - ma - b\sqrt {1 + {m^2}} }}{{\sqrt {1 + {m^2}} }}} \right| = b$

Cross multiplying the terms, we get,

$- ma - b\sqrt {1 + {m^2}} = \pm b\sqrt {1 + {m^2}}$

If we take negative signs into consideration, we get $m = 0$. But we have to c=find the positive value of m. So, this is not the correct answer.

If we take positive sign, we get,

$- ma = 2b\sqrt {1 + {m^2}}$

Squaring both the sides of the equation, we get,

${\left( { - ma} \right)^2} = {\left( {2b\sqrt {1 + {m^2}} } \right)^2}$

$\Rightarrow {m^2}{a^2} = 4{b^2}\left( {1 + {m^2}} \right)$

Opening bracket and simplifying the expression, we get,

${m^2}{a^2} = 4{b^2} + 4{b^2}{m^2}$

Shifting all the terms consisting m to left side of the equation, we get,

${m^2}{a^2} - 4{b^2}{m^2} = 4{b^2}$

$\Rightarrow {m^2} = \dfrac{{4{b^2}}}{{\left( {{a^2} - 4{b^2}} \right)}}$

Calculating the value of m, we get,

$\therefore m = \sqrt {\dfrac{{4{b^2}}}{{\left( {{a^2} - 4{b^2}} \right)}}} = \dfrac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}$

So, the positive value of m for which $y = mx - b\sqrt {1 + {m^2}}$ is a common tangent to ${x^2} + {y^2} = {b^2}$ and ${\left( {x - a} \right)^2} + {y^2} = {b^2}$ is $\dfrac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}$.

Hence, option A is correct.

Note: The given question also involves basic understanding of equations of conic sections. Such problems illustrate the interdependence of mathematical ideas and topics on each other. Care should be taken while handling the calculative steps so as to be sure of the final answer for the slope of common tangent. Such problems put our knowledge of all the fields of mathematics such as algebra, calculus, conic sections and analytical geometry to test.