Q.A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $x = \cos 2t + \log(\tan t)$ and $y = 2t + \cot 2t$, then $\dfrac{dy}{dx} =$ (A) $\tan 2t$ (B) $-\csc 2t$ (C) $-\cot 2t$ (D) $\sec 2t$

   ›Reveal solutionSolution

   Differentiating parametrically, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=-\csc 2t$.

   Differentiate $x$ w.r.t. $t$.

   $\frac{dx}{dt}=-2\sin 2t+\frac{\sec^2t}{\tan t}=-2\sin 2t+\frac{1}{\sin t\cos t}=-2\sin 2t+2\csc 2t=\frac{2\cos^2 2t}{\sin 2t}.$

   Differentiate $y$ w.r.t. $t$.

   $\frac{dy}{dt}=2-2\csc^2 2t=2\cdot\frac{\sin^2 2t-1}{\sin^2 2t}=\frac{-2\cos^2 2t}{\sin^2 2t}.$

   Divide. …

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2. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.The ratio of the length of the subnormal to the square of the length of the subtangent at any point P on the curve $y^2 = (2x+1)^3$ is (A) $27$ (B) $\dfrac{1}{9}$ (C) $9$ (D) $\dfrac{8}{27}$

   ›Reveal solutionSolution

   Subnormal $=y\,y'$, subtangent $=y/y'$; the required ratio is $y'^3/y = 27$.

   For any curve, the subnormal has length $\left|y\dfrac{dy}{dx}\right|$ and the subtangent has length $\left|\dfrac{y}{dy/dx}\right|$. Hence

   $\frac{\text{subnormal}}{(\text{subtangent})^2}=\frac{y\,y'}{\left(y/y'\right)^2}=\frac{y'^{\,3}}{y}.$

   Differentiate $y^2=(2x+1)^3$:

   $2y\,y'=3(2x+1)^2\cdot 2=6(2x+1)^2\;\Rightarrow\; y'=\frac{3(2x+1)^2}{y}.$

   With $y=(2x+1)^{3/2}$, …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the rate of change of volume of a cube and that of its surface area are numerically equal, then the length of its diagonal is (A) $2\sqrt{3}$ (B) $\sqrt{3}$ (C) $4\sqrt{3}$ (D) $6\sqrt{3}$

   ›Reveal solutionSolution

   Equating $\dfrac{dV}{dt}$ and $\dfrac{dS}{dt}$ numerically gives edge $x=4$, so the diagonal is $4\sqrt{3}$.

   Let the edge length be $x$. Then

   $V = x^3 \implies \frac{dV}{dt} = 3x^2\,\frac{dx}{dt},$

   $S = 6x^2 \implies \frac{dS}{dt} = 12x\,\frac{dx}{dt}.$

   Numerically equal:

   $3x^2 = 12x \implies x = 4.$ …

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4. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The height of a cone with semi vertical angle $\dfrac{\pi}{3}$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\sqrt{3}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   The problem uses related rates with the cone’s volume fixed. Differentiating $V = \frac13 \pi r^2 h$ and using $\frac{dh}{dt}=2$ gives $\frac{dr}{dt} = -\frac{r}{2h} \cdot 2$. With semi-vertical angle $\pi/3$, $\frac{r}{h} = \tan(\pi/3)=\sqrt{3}$, so $\frac{dr}{dt} = -\sqrt{3}$ units/min. The rate of decrease is $\sqrt{3}$.

   Concept & Intuition

   We have a cone whose height is increasing, but we want its volume to stay constant. That means the radius must shrink to compensate. The key is to relate the radius and height through the fixed semi-vertical angle — this gives a constant ratio $r/h = \tan(\pi/3) = \sqrt{3}$. Then we use calculus (related rates) to find how fast the radius must change when the height changes at 2 units/min.

   Step-by-step solution

   1. Volume of a cone

      The volume is $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant, so $\frac{dV}{dt} = 0$.

   2. Differentiate implicitly with respect to time

      Using the product rule:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (nonzero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   3. Solve for $\frac{dr}{dt}$

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   4. Use the given rate and geometry We are told $\frac{dh}{dt} = 2$ units/min. The semi-vertical angle is $\pi/3$, so in a right triangle formed by the height, radius, and slant height: $$$$

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5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The height of a cone with semi vertical angle $\pi/3$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   For a cone of fixed volume, the radius must shrink at a rate that exactly compensates the growth in height. Using the relation $V = \frac13 \pi r^2 h$ and differentiating with respect to time gives $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$. With semi-vertical angle $\pi/3$, we have $r/h = \tan(\pi/3) = \sqrt{3}$, so $\frac{dr}{dt} = -\frac{\sqrt{3}}{2} \cdot 2 = -\sqrt{3}$ units/min. The required rate of decrease is $\sqrt{3}$ units/min, so the correct option is (A).

   The key idea is that "fixed volume" ties the radius and height together through a constraint. When one changes, the other must change in a specific way to keep the product $r^2 h$ constant. The semi-vertical angle gives the instantaneous ratio of radius to height at the moment we are considering — that ratio is not constant over time (since the cone's shape changes), but at the instant we care about, it is fixed by the given angle.

   Let’s work through it step by step.

   1. Write the volume constraint. For a cone, $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant. Differentiating both sides with respect to time $t$:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (non-zero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   2. Solve for $\frac{dr}{dt}$. Rearranging:

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt}.$

   Assuming $r \neq 0$, divide both sides by $r$:

   $2h \frac{dr}{dt} = - r \frac{dh}{dt}.$

   Hence:

   $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   The negative sign tells us that if height increases, radius must decrease — exactly what we expect.

   3. Use the semi-vertical angle to find $r/h$. The semi-vertical angle is the angle between the axis and the slant height. In a right circular cone, $\tan(\text{semi-vertical angle}) = \frac{\text{radius}}{\text{height}}$. Given the angle is $\pi/3$:

   $\frac{r}{h} = \tan\frac{\pi}{3} = \sqrt{3}.$

   So $r = \sqrt{3}\,h$ at the instant under consideration.

   4. Plug in the given rate. …

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6. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The equation of the normal drawn to the curve $y = \sin 3x$ at $x = \frac{\pi}{4}$ is (A) $y = \frac{\sqrt{3}}{2} \left( x + \frac{6 - \pi}{4} \right)$ (B) $y = \frac{\sqrt{2}}{3} \left( x + \frac{6 - \pi}{4} \right)$ (C) $y = \frac{\sqrt{3}}{2} \left( x - \frac{6 - \pi}{4} \right)$ (D) $y = \frac{\sqrt{2}}{3} \left( x - \frac{6 - \pi}{4} \right)$

   ›Reveal solutionSolution

   The normal line is perpendicular to the tangent. We find the slope of the tangent via differentiation, take its negative reciprocal, then use the point-slope form with the point on the curve at $x = \pi/4$. The correct equation is option (D).

   The key idea: a normal line is just the line through a point on a curve whose slope is the negative reciprocal of the derivative at that point. So we need two things — the coordinates of the point, and the slope of the tangent there.

   1. Find the point on the curve. At $x = \frac{\pi}{4}$,

   $y = \sin\left(3 \cdot \frac{\pi}{4}\right) = \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}.$

   So the point is $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$.

   2. Find the slope of the tangent. Differentiate:

   $\frac{dy}{dx} = 3\cos 3x.$

   At $x = \frac{\pi}{4}$,

   $\frac{dy}{dx}\Big|_{x=\pi/4} = 3\cos\frac{3\pi}{4} = 3\left(-\frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2}.$

   That’s the slope of the tangent.

   3. Slope of the normal. The normal is perpendicular, so its slope $m$ satisfies

   $m \cdot \left(-\frac{3\sqrt{2}}{2}\right) = -1 \quad\Rightarrow\quad m = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}.$

   4. Equation of the normal. Using point-slope form:

   $y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{3}\left(x - \frac{\pi}{4}\right).$

   Multiply through:

   $y = \frac{\sqrt{2}}{3}x - \frac{\sqrt{2}}{3}\cdot\frac{\pi}{4} + \frac{\sqrt{2}}{2}.$

   Combine the constant terms: …

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7. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.A rod of length 6 units slides with its ends on the coordinate axes. The locus of the midpoint of the rod is (A) $x^2 + y^2 = 9$ (B) $x + y = 3$ (C) $x^2 + y^2 = 36$ (D) $x + y = 6$

   ›Reveal solutionSolution

   The midpoint of a rod sliding with its ends on the axes always lies on a circle centered at the origin with radius half the rod’s length. Here, the locus is $x^2 + y^2 = 9$.

   The key idea is that the rod’s ends are constrained to the $x$-axis and $y$-axis. As the rod slides, its midpoint traces a curve. We need to find the equation that describes all possible positions of that midpoint.

   Think about it: if you fix the rod’s length, the distance from the midpoint to each end is constant. But more importantly, the midpoint’s coordinates are simply the averages of the coordinates of the ends. Since the ends lie on the axes, their coordinates are $(a,0)$ and $(0,b)$, where $a$ and $b$ can vary but the distance between them is fixed at 6.

   The distance condition gives a relation between $a$ and $b$. The midpoint’s coordinates are expressed in terms of $a$ and $b$. Eliminating $a$ and $b$ then yields the locus.

   Let’s work through it step by step.

   1. Let the ends of the rod be at $A(a,0)$ on the $x$-axis and $B(0,b)$ on the $y$-axis. The rod’s length is 6, so the distance $AB = 6$.

   2. Using the distance formula:

   $\sqrt{(a-0)^2 + (0-b)^2} = 6$

   Squaring both sides:

   $a^2 + b^2 = 36$

   This is the constraint linking $a$ and $b$.

   3. The midpoint $M$ of the rod has coordinates:

   $M = \left( \frac{a+0}{2}, \frac{0+b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right)$

   Let $M = (x, y)$. Then:

   $x = \frac{a}{2}, \quad y = \frac{b}{2}$

   So $a = 2x$ and $b = 2y$. …

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8. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $\sin y = \sin 3t$ and $x = \sin t$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{3}{\sqrt{4-x^2}}$ (B) $\dfrac{3}{\sqrt{1-x^2}}$ (C) $\dfrac{1}{\sqrt{4-x^2}}$ (D) $\dfrac{-1}{\sqrt{4-x^2}}$

   ›Reveal solutionSolution

   Treat both $y$ and $x$ as functions of $t$ (parametric differentiation): $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. With $y=3t$ and $x=\sin t$, this gives $\dfrac{3}{\sqrt{1-x^2}}$ — option (B).

   Concept. We are not given $y$ as a function of $x$ directly; instead both are tied to the parameter $t$. Parametric differentiation says $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ whenever $dx/dt\neq 0$.

   Step 1 — read off $y$ in terms of $t$.

   The relation $\sin y=\sin 3t$ has the principal solution $y=3t$, so

   $\frac{dy}{dt}=3.$

   Step 2 — differentiate $x=\sin t$.

   $\frac{dx}{dt}=\cos t.$

   Step 3 — form the ratio. …

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9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The slope of the tangent drawn from the point (1,1) to the hyperbola $2x^2 - y^2 = 4$ is (A) $2$ (B) $\dfrac{-2 \pm \sqrt{6}}{2}$ (C) $-1 \pm \sqrt{6}$ (D) $\dfrac{-2 \pm \sqrt{3}}{2}$

   ›Reveal solutionSolution

   The key idea is to write the equation of a tangent to the hyperbola in slope form, then impose that it passes through (1,1). Solving the resulting quadratic gives the slopes, which match option (B).

   We are given the hyperbola $2x^2 - y^2 = 4$. First, rewrite it in standard form:

   $\frac{x^2}{2} - \frac{y^2}{4} = 1$

   So $a^2 = 2$ and $b^2 = 4$.

   ---

   Concept & Intuition

   For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, any non-vertical tangent line can be written in the slope form:

   $y = mx \pm \sqrt{a^2 m^2 - b^2}$

   This formula comes from solving the condition that the line $y = mx + c$ touches the hyperbola (i.e., the quadratic in $x$ has a double root). The $\pm$ accounts for two parallel tangents with the same slope.

   Here, we want tangents that also pass through the external point $(1,1)$. So we substitute $x=1, y=1$ into the tangent equation and solve for $m$.

   ---

   Step-by-step solution

   1. Write the tangent equation in slope form For $\frac{x^2}{2} - \frac{y^2}{4} = 1$, we have $a^2 = 2$, $b^2 = 4$. The tangent with slope $m$ is:

   $y = mx \pm \sqrt{2m^2 - 4}$

   2. Impose that the point $(1,1)$ lies on this line Substitute $x = 1$, $y = 1$:

   $1 = m(1) \pm \sqrt{2m^2 - 4}$

   So:

   $1 - m = \pm \sqrt{2m^2 - 4}$

   3. Square both sides (but remember this may introduce extraneous solutions — we’ll check later)

   $(1 - m)^2 = 2m^2 - 4$

   Expand:

   $1 - 2m + m^2 = 2m^2 - 4$

   4. Rearrange into a quadratic

   $0 = 2m^2 - 4 - m^2 + 2m - 1$

   $0 = m^2 + 2m - 5$

   So:

   $m^2 + 2m - 5 = 0$

   5. Solve the quadratic

   $m = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2}$

   Simplify:

   $m = -1 \pm \sqrt{6}$

   6. Check for extraneous solutions The original equation required $1 - m = \pm \sqrt{2m^2 - 4}$. For $m = -1 + \sqrt{6} \approx 1.449$: $1 - m \approx -0.449$ (negative), so we need the negative square root: $\sqrt{2m^2 - 4} \approx \sqrt{2(2.1) - 4} = \sqrt{0.2} \approx 0.447$ — the negative of that is $-0.447$, which matches. For $m = -1 - \sqrt{6} \approx -3.449$: $1 - m \approx 4.449$ (positive), and $\sqrt{2m^2 - 4}$ is large positive — so the positive sign works. Both are valid.

   Watch out

   A common mistake is to stop at $m = -1 \pm \sqrt{6}$ and pick option (C). But look carefully: the problem asks for the slope of the tangent drawn from the point (1,1). The slopes we found are indeed $-1 \pm \sqrt{6}$, but that is not one of the given options! Option (C) is $-1 \pm \sqrt{6}$, which matches our result — so why isn’t it correct? …

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10. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.If the base of an isosceles triangle is $3\sqrt{2}$ feet and the two equal sides of it are increasing at the rate of $1$ ft/s, then the rate of increase of its area (in sq.ft/sec) when the angle between the equal sides is a right angle is (A) $3\sqrt{3}$ (B) $\sqrt{3}$ (C) $9$ (D) $3$

    ›Reveal solutionSolution

    The area of an isosceles triangle is expressed in terms of the equal side length and the included angle. Using the given rate of change of the side and the fact that the angle is fixed at the instant of interest, the rate of increase of area is found to be $3$ sq.ft/sec.

    The problem gives an isosceles triangle with base $3\sqrt{2}$ feet and equal sides that are increasing at $1$ ft/s. We need the rate of increase of its area at the moment when the angle between the equal sides is a right angle.

    The key is to choose a formula for area that directly involves the changing quantity (the equal side length) and the angle. For any triangle, area is $\frac12 ab \sin C$. Here, the two equal sides are the ones forming the included angle, so that formula is perfect.

    1. Set up the variables. Let the equal sides each have length $s$ feet, and let $\theta$ be the angle between them. The area $A$ of the triangle is

    $A = \frac12 \cdot s \cdot s \cdot \sin\theta = \frac12 s^2 \sin\theta.$

    2. What is given and what is wanted? We know $\frac{ds}{dt} = 1$ ft/s. We want $\frac{dA}{dt}$ at the instant when $\theta = 90^\circ = \frac{\pi}{2}$ radians. But note: the base is fixed at $3\sqrt{2}$ feet. Does that give a relation between $s$ and $\theta$? Yes — by the law of cosines, the base $b$ satisfies

    $b^2 = s^2 + s^2 - 2s^2 \cos\theta = 2s^2(1 - \cos\theta).$

    So $b = 3\sqrt{2}$ is constant, meaning $s$ and $\theta$ are not independent — as $s$ increases, $\theta$ must change to keep the base fixed. However, we only need the rate at a specific instant, not a full functional relation.

    3. Differentiate the area with respect to time. Since both $s$ and $\theta$ can change with time,

    $\frac{dA}{dt} = \frac12 \left( 2s \frac{ds}{dt} \sin\theta + s^2 \cos\theta \cdot \frac{d\theta}{dt} \right) = s \frac{ds}{dt} \sin\theta + \frac12 s^2 \cos\theta \frac{d\theta}{dt}.$

    4. Find $s$ and $\frac{d\theta}{dt}$ at the required instant. At $\theta = \frac{\pi}{2}$, $\sin\theta = 1$, $\cos\theta = 0$. The law of cosines gives

    $(3\sqrt{2})^2 = 2s^2 (1 - \cos\frac{\pi}{2}) = 2s^2 (1 - 0) = 2s^2.$

    So $18 = 2s^2$, hence $s^2 = 9$ and $s = 3$ feet (positive length).

    Now we need $\frac{d\theta}{dt}$ at that instant. Differentiate the law of cosines relation with respect to time. From $b^2 = 2s^2(1 - \cos\theta)$, since $b$ is constant,

    $0 = \frac{d}{dt}\left[2s^2(1 - \cos\theta)\right] = 4s \frac{ds}{dt}(1 - \cos\theta) + 2s^2 \sin\theta \frac{d\theta}{dt}.$

    At $\theta = \frac{\pi}{2}$, $\cos\theta = 0$, $\sin\theta = 1$, $s = 3$, $\frac{ds}{dt} = 1$: …

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11. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.If P is any point on the curve $y^2 = 4ax$, other than the origin, then the length of the subtangent at P, y-coordinate of P and the length of the subnormal at P are in (A) arithmetic progression (B) arithmetic-Geometric progression (C) harmonic progression (D) geometric progression

    ›Reveal solutionSolution

    For a point $P$ on the parabola $y^2 = 4ax$, the subtangent, the y-coordinate, and the subnormal are in geometric progression. The correct option is (D).

    ---

    The key idea is to recall the geometric meanings of subtangent and subnormal for a curve. For a point $P$ on a curve, the subtangent is the projection of the tangent segment onto the x-axis, and the subnormal is the projection of the normal segment onto the x-axis. Their lengths have simple formulas in terms of the derivative.

    For any curve $y = f(x)$, at a point $(x, y)$:

    • Length of subtangent = $\left| \frac{y}{dy/dx} \right|$
    • Length of subnormal = $\left| y \cdot \frac{dy}{dx} \right|$

    Here the curve is given implicitly as $y^2 = 4ax$. We’ll use these formulas and then check the progression among the three quantities.

    ---

    1. Find the derivative. Differentiate $y^2 = 4ax$ with respect to $x$:

    $2y \frac{dy}{dx} = 4a \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2a}{y}.$

    2. Compute the subtangent.

       Subtangent length = $\left| \frac{y}{dy/dx} \right| = \left| \frac{y}{2a/y} \right| = \left| \frac{y^2}{2a} \right|$.

       Since $y^2 = 4ax$, this becomes $\left| \frac{4ax}{2a} \right| = 2|x|$.

       For a point on the parabola (other than the origin), $x > 0$ (right-opening parabola), so subtangent = $2x$.

    3. The y-coordinate of P is simply $y$.

    4. Compute the subnormal.

       Subnormal length = $\left| y \cdot \frac{dy}{dx} \right| = \left| y \cdot \frac{2a}{y} \right| = 2a$.

       Notice this is constant — independent of the point $P$! That’s a neat property of the parabola.

    5. Now we have three numbers:

       • Subtangent: $2x$
       • y-coordinate: $y$
       • Subnormal: $2a$

       We need to check if they are in arithmetic, geometric, or harmonic progression. …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The slope of a common tangent to the circles $x^2 + y^2 = 16$ and $(x-9)^2 + y^2 = 16$ is (A) $\dfrac{8}{\sqrt{13}}$ (B) $\dfrac{4}{\sqrt{13}}$ (C) $\dfrac{\sqrt{17}}{8}$ (D) $\dfrac{8}{\sqrt{17}}$

    ›Reveal solutionSolution

    The common tangent to two equal circles is found by considering the line’s distance from each centre equals the radius; solving gives slope $\pm \frac{8}{\sqrt{17}}$, so the correct option is (D).

    We have two circles of equal radius $4$:

    • Circle 1: centre $O_1 = (0,0)$, radius $r = 4$
    • Circle 2: centre $O_2 = (9,0)$, radius $r = 4$

    A common tangent touches both circles. Because the circles are the same size, the common tangents are either direct (parallel to the line joining centres) or transverse (crossing between them). Here we want a common tangent — the slope will be the same for both points of tangency.

    ---

    Why this approach works

    For any line to be tangent to a circle, the perpendicular distance from the circle’s centre to the line must equal the radius. If the same line is tangent to both circles, then the distances from $O_1$ and $O_2$ to the line are both $4$. This gives two equations in the line’s parameters, which we can solve for the slope.

    ---

    Step-by-step solution

    1. Write the general line equation Let the common tangent have slope $m$ and intercept $c$:

    $y = mx + c \quad \Rightarrow \quad mx - y + c = 0$

    2. Distance from centre $(0,0)$ to the line equals radius $4$

    $\frac{|m\cdot 0 - 0 + c|}{\sqrt{m^2 + 1}} = 4 \quad \Rightarrow \quad \frac{|c|}{\sqrt{m^2+1}} = 4 \qquad(1)$

    3. Distance from centre $(9,0)$ to the same line also equals $4$

    $\frac{|m\cdot 9 - 0 + c|}{\sqrt{m^2+1}} = 4 \quad \Rightarrow \quad \frac{|9m + c|}{\sqrt{m^2+1}} = 4 \qquad(2)$

    4. Equate the two distances From (1) and (2):

    $|c| = |9m + c|$

    This gives two cases:

    • Case 1: $c = 9m + c \quad \Rightarrow \quad 9m = 0 \quad \Rightarrow \quad m = 0$

      Then from (1): $\frac{|c|}{1} = 4 \Rightarrow c = \pm 4$.

      This gives horizontal tangents $y = \pm 4$ — these are indeed common tangents (top and bottom), but slope $0$ is not among the options.

    • Case 2: $c = -(9m + c) \quad \Rightarrow \quad 2c = -9m \quad \Rightarrow \quad c = -\frac{9m}{2}$

    5. Substitute $c$ into the distance condition (1) …

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