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**Hint**: To solve this question we will assume the coordinates of the centre of the circle and the radius of the circle then will find the angle of the trigonometric function and then the slope of the equation.

**:**

__Complete step-by-step answer__Let the centre of the circle be $ (h,r) $ as it touches the x-axis.

Therefore, the equation of circle is –

$ {(x - h)^2} + {(y - r)^2} = {r^2} $ .... (A)

The circle also touches at $ y = mx $

Let m be the slope which can be expressed as –

$ m = \tan \theta $ .... (B) (Where the angle of the function is made with x-axis)

$ \therefore r = h\tan \left( {\dfrac{\theta }{2}} \right) $

Tangent is the inverse function of cot. So, place $ \tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{1}{{\cot \left( {\dfrac{\theta }{2}} \right)}} $ in the above equation.

$ \therefore r = h\left( {\dfrac{1}{{\cot \left( {\dfrac{\theta }{2}} \right)}}} \right) $

Do cross multiplication, Where numerator of one side is multiplied with the denominator of the opposite side.

$ \Rightarrow r.\cot \left( {\dfrac{\theta }{2}} \right) = h $

The above equation can be re-written as –

$ \Rightarrow h = r.\cot \left( {\dfrac{\theta }{2}} \right) $ .... (C)

Place the above value in equation (A)

$ {(x - r.\cot \left( {\dfrac{\theta }{2}} \right))^2} + {(y - r)^2} = {r^2} $

Since both the circles passes through the points $ (6,4) $

$ \Rightarrow {(6 - r.\cot \left( {\dfrac{\theta }{2}} \right))^2} + {(4 - r)^2} = {r^2} $

Simplify the above equation using the identity $ {(a - b)^2} = {a^2} - 2ab + {b^2} $

$ \Rightarrow 36 - {r^2}.{\cot ^2}\left( {\dfrac{\theta }{2}} \right) - 12r.\cot \left( {\dfrac{\theta }{2}} \right) + 16{r^2} - 8r = {r^2} $

Simplify the above equation –

$ \Rightarrow {r^2}.{\cot ^2}\left( {\dfrac{\theta }{2}} \right) + r\left( { - 12r.\cot \left( {\dfrac{\theta }{2}} \right) - 8} \right) + 52 = 0 $

Product of roots is given by $ \dfrac{{52}}{3} $

We know that in quadratic equation –

Product of roots is given by $ \dfrac{c}{a} $

Now,

$ \dfrac{c}{a} = \dfrac{{52}}{3} $

Here we have $ a = {\cot ^2}\left( {\dfrac{\theta }{2}} \right) $ and $ c = 52 $ place values in the above equation –

$ \Rightarrow \dfrac{{52}}{{{{\cot }^2}\left( {\dfrac{\theta }{2}} \right)}} = \dfrac{{52}}{3} $

Common multiples from both the sides of the equation cancel each other.

$ \Rightarrow \dfrac{1}{{{{\cot }^2}\left( {\dfrac{\theta }{2}} \right)}} = \dfrac{1}{3} $

The above equation implies –

$ \Rightarrow {\cot ^2}\left( {\dfrac{\theta }{2}} \right) = 3 $

Cot is the inverse function of tangent –

$ \Rightarrow \dfrac{1}{{{{\tan }^2}\left( {\dfrac{\theta }{2}} \right)}} = 3 $

Cross-multiplication implies –

$ \Rightarrow {\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{1}{3} $

Take square-root on both the side of the equation –

$ \Rightarrow \tan \left( {\dfrac{\theta }{2}} \right) = \pm \dfrac{1}{{\sqrt 3 }} $

Since the circle is in the first quadrant, the tangent is positive.

$ \Rightarrow \tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{1}{{\sqrt 3 }} $

By referring to the trigonometric table for values –

$ \Rightarrow \tan \left( {\dfrac{\theta }{2}} \right) = \tan \dfrac{\pi }{6} $

Tangent on both the sides of the equation cancels each other.

$ \Rightarrow \left( {\dfrac{\theta }{2}} \right) = \dfrac{\pi }{6} $

Simplify the above equation-

$ \Rightarrow \theta = \dfrac{{\pi \times 2}}{6} $

Common factors from the numerator and the denominator cancel each other.

$ \Rightarrow \theta = \dfrac{\pi }{3} $

Place the above value in equation (B)

$

\tan \theta = \sqrt 3 \\

\Rightarrow m = \sqrt 3 \;

$

This is the required solution.

**So, the correct answer is “ $ m = \sqrt 3 $ ”.**

**Note**: Follow the All STC rule, and first check for which quadrant the point lies and then take positive or the negative signs for the trigonometric functions. It is most important to follow the correct sign convention.

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