Question

If \[a>2b>0\] then the positive values of m for which \[y=mx-b\sqrt{1+{{m}^{2}}}\] is a common tangent to \[{{x}^{2}}+{{y}^{2}}={{b}^{2}}\] and \[{{(x-a)}^{2}}+{{y}^{2}}={{b}^{2}}\] is
(1) \[\dfrac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
(2) \[\dfrac{\sqrt{{{a}^{2}}-4{{b}^{2}}}}{2b}\]
(3) \[\dfrac{2b}{(a-2b)}\]
(4) \[\dfrac{b}{(a-2b)}\]

Answer 1

Hint: When the line touches the circle then it is known as the tangent to the circle and there is the perpendicular distance between the point where the tangent touches the circle and the center of the circle and that perpendicular distance is also called the radius. All equations of the tangent to the circle should be clear to you.

Complete step-by-step solution:
In the above question, it is given that \[{{x}^{2}}+{{y}^{2}}={{b}^{2}}\] and \[{{(x-a)}^{2}}+{{y}^{2}}={{b}^{2}}\] are the circles and \[y=mx-b\sqrt{1+{{m}^{2}}}\] is the common tangent to both the circles and we have to find all the possible values of m for which the above tangent is common to both the circles. So it is as follows.
In the first circle \[{{x}^{2}}+{{y}^{2}}={{b}^{2}}\], tangent will be \[y=mx-b\sqrt{1+{{m}^{2}}}\] for all the possible values of m. where m is the slope of the tangent
Now we will find the circle \[{{(x-a)}^{2}}+{{y}^{2}}={{b}^{2}}\]. In this circle, the center of this circle is \[(a,0)\] and the radius of this circle is b.
We know that the distance between tangent and center will be the radius of the circle and the perpendicular distance between any two points is given by the formula below
\[P=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
Where P is the perpendicular distance between the points, a and b are the coefficients of x and y in the equation while \[({{x}_{1}},{{y}_{1}})\] will be the points. So now we will solve our question further. So the perpendicular distance will be given by-
\[\pm b=\dfrac{ma-b\sqrt{1+{{m}^{2}}}}{\sqrt{1+{{m}^{2}}}}\]
After removing the modulus operator the value of b becomes both negative and positive so we will solve them separately.
First, we will solve for the negative value of b, which is as shown below.
\[-b=\dfrac{ma-b\sqrt{1+{{m}^{2}}}}{\sqrt{1+{{m}^{2}}}}\]
\[\Rightarrow -b\sqrt{1+{{m}^{2}}}=ma-b\sqrt{1+{{m}^{2}}}\]
\[ma=0\], which gives the value of m equals zero. As slope cannot be equal to zero so this value does not hold.
Now we will solve for the positive value of b, which gives
\[+b=\dfrac{ma-b\sqrt{1+{{m}^{2}}}}{\sqrt{1+{{m}^{2}}}}\]
\[\begin{align}
  & \Rightarrow b\sqrt{1+{{m}^{2}}}=ma-b\sqrt{1+{{m}^{2}}} \\
 & \Rightarrow b\sqrt{1+{{m}^{2}}}+b\sqrt{1+{{m}^{2}}}=ma \\
 & \Rightarrow 2b\sqrt{1+{{m}^{2}}}=ma \\
\end{align}\]
On squaring both the sides we get,
\[\begin{align}
  & {{(2b\sqrt{1+{{m}^{2}}})}^{2}}={{m}^{2}}{{a}^{2}} \\
 & \Rightarrow 4{{b}^{2}}(1+{{m}^{2}})={{m}^{2}}{{a}^{2}} \\
 & \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 4{{b}^{2}}+4{{b}^{2}}{{m}^{2}}={{m}^{2}}{{a}^{2}} \\
 & \Rightarrow 4{{b}^{2}}={{m}^{2}}{{a}^{2}}-4{{b}^{2}}{{m}^{2}} \\
 & \Rightarrow 4{{b}^{2}}={{m}^{2}}({{a}^{2}}-4{{b}^{2}}) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{m}^{2}}=\frac{4{{b}^{2}}}{{{a}^{2}}-4{{b}^{2}}} \\
 & \Rightarrow m=\dfrac{\sqrt{4{{b}^{2}}}}{\sqrt{{{a}^{2}}-4{{b}^{2}}}} \\
\end{align}\]
\[\Rightarrow m=\dfrac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]
So for the value of \[m=\dfrac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\] , the tangent is common to both the circles.
So the correct answer is (1) \[m=\dfrac{2b}{\sqrt{{{a}^{2}}-4{{b}^{2}}}}\]

Note:Tangent at the circle is always perpendicular to the radius of the circle. The point where the tangent touches the circle is known as the point of tangency. Also, the tangent always touches the circle but never crosses the circle. If two tangents come from the same point to the circle then both the tangents will be equal.