Given hyperbola and point of contact

The hyperbola is $H: x^2-y^2=1$ and the touching point is $P(x_1,y_1), \, x_1>1,\; y_1>0.$

Since $P$ lies on the hyperbola, $x_1^2-y_1^2=1 \implies y_1=\sqrt{x_1^2-1}.$

Equation of tangent to the hyperbola at $P$

For $x^2-y^2=1$, the tangent at $(x_1,y_1)$ is $x x_1 - y y_1 = 1.$

This tangent meets the $x$-axis at $M$. On the $x$-axis, $y=0$, so $x x_1=1 \implies x=\frac1{x_1}.$ Hence, $M\left(\frac1{x_1},0\right).$

Finding the center $N$ of the circle

The circle has center $N(x_2,0)$ and is tangent to the hyperbola at $P$.

Since the tangent is common to both curves at $P$, the radius $NP$ is perpendicular to the tangent.

Slope of tangent to the hyperbola: Differentiate $x^2-y^2=1$: $2x-2y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=\frac{x}{y}.$ So at $P$ the tangent slope is $\frac{x_1}{y_1}.$

Therefore slope of radius $NP$ is $-\frac{y_1}{x_1}.$

But slope of line joining $N(x_2,0)$ and $P(x_1,y_1)$ is $\frac{y_1-0}{x_1-x_2}=\frac{y_1}{x_1-x_2}.$

Equating, $\frac{y_1}{x_1-x_2}=-\frac{y_1}{x_1}.$ Since $y_1>0$, cancel $y_1$: $\frac{1}{x_1-x_2}=-\frac{1}{x_1}$ $x_1=-(x_1-x_2)$ $x_2=2x_1.$

Thus, $N(2x_1,0).$

Coordinates of centroid $(l,m)$ of triangle $PMN$

The vertices are $P(x_1,y_1),\quad M\left(\frac1{x_1},0\right),\quad N(2x_1,0).$

So centroid is $l=\frac{x_1+\frac1{x_1}+2x_1}{3}=x_1+\frac{1}{3x_1},$ $m=\frac{y_1+0+0}{3}=\frac{y_1}{3}.$

Check option A and C: $\dfrac{dl}{dx_1}$

Differentiate $l=x_1+\frac{1}{3x_1}.$

Then $\frac{dl}{dx_1}=1-\frac{1}{3x_1^2}.$

Therefore:

• A is correct
• C is incorrect

Check option B: $\dfrac{dm}{dx_1}$

Since $m=\frac{y_1}{3}=\frac{1}{3}\sqrt{x_1^2-1},$ differentiate w.r.t. $x_1$:

Check option D: $\dfrac{dm}{dy_1}$

Since $m=\frac{y_1}{3},$ directly, $\frac{dm}{dy_1}=\frac13.$

Hence D is correct.

Final selection

Correct options are $\boxed{A,\;B,\;D}$

Comparison with stored answer

Stored correct answer: $A,B,D$

This matches exactly.