Q.The solution of the differential equation 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 represents a family of
(A) straight lines
(B) parabolas
(C) Circles
(D) Ellipses

Showing the 12 most recent of 153 on this concept.

1. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.A common tangent to the circle $x^2+y^2=9$ and parabola $y^2=8x$ is (A) $3x-\sqrt{3}y+2=0$ (B) $x-\sqrt{3}y+6=0$ (C) $2x-\sqrt{3}y+3=0$ (D) $x-3y+6=0$

   ›Reveal solutionSolution

   Write the parabola's tangent in slope form, force it to also be tangent to the circle (distance from centre = radius), solve for the slope, then verify.

   Concept and Intuition

   The tangent to $y^2=4ax$ with slope $m$ is $y=mx+a/m$. A line is tangent to a circle centred at the origin with radius $r$ exactly when its perpendicular distance from the origin equals $r$.

   Step-by-Step Solution

   1. Parabola $y^2=8x$: $4a=8\Rightarrow a=2$. Tangent: $y=mx+2/m$, i.e. $mx-y+2/m=0$.
   2. Distance from origin (circle centre, radius 3): $\dfrac{|2/m|}{\sqrt{m^2+1}}=3\Rightarrow \dfrac{4}{m^2}=9(m^2+1)\Rightarrow 9m^4+9m^2-4=0$.
   3. Let $u=m^2$: $9u^2+9u-4=0\Rightarrow u=\dfrac{-9\pm\sqrt{81+144}}{18}=\dfrac{-9\pm15}{18}$, giving $u=1/3$ (rejecting the negative root).
   4. So $m=\pm1/\sqrt3$. Take $m=1/\sqrt3$: $2/m=2\sqrt3$. Tangent: $y=\dfrac{x}{\sqrt3}+2\sqrt3\Rightarrow \sqrt3y=x+6\Rightarrow x-\sqrt3y+6=0$. …

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2. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Parametric equations of the circle $2x^2+2y^2=9$ are (A) $x=\frac{3}{2}\cos\theta,\ y=\frac{3}{2}\sin\theta$ (B) $x=\frac{3}{\sqrt{2}}\cos\theta,\ y=3\sin\theta$ (C) $x=\frac{3}{\sqrt{2}}\sin\theta,\ y=\frac{3}{\sqrt{2}}\cos\theta$ (D) $x=3\sin\theta,\ y=\frac{3}{2}\cos\theta$

   ›Reveal solutionSolution

   The circle has radius $\frac{3}{\sqrt2}$, and only option (C) uses that exact radius consistently for both coordinates (with sin/cos swapped, still a valid circle parametrization).

   Concept and Intuition

   A circle $x^2+y^2=r^2$ has parametric form $x=r\cos\theta,\ y=r\sin\theta$ (or any rotation/reflection of it, e.g. swapping sine and cosine) — the key requirement is that both coordinates carry the same radius $r$.

   Step-by-Step Solution

   1. $2x^2+2y^2=9 \Rightarrow x^2+y^2=\dfrac92$, so $r=\sqrt{9/2}=\dfrac{3}{\sqrt2}$.
   2. Check (A): coefficient $\frac32=1.5 \ne \frac{3}{\sqrt2}\approx2.12$ — wrong radius.
   3. Check (B): $x$ has coefficient $\frac{3}{\sqrt2}$ but $y$ has coefficient $3$ — inconsistent, not a circle of this radius (fails $x^2+y^2=r^2$ identity).
   4. Check (C): $x=\frac{3}{\sqrt2}\sin\theta,\ y=\frac{3}{\sqrt2}\cos\theta$: $x^2+y^2=\frac92(\sin^2\theta+\cos^2\theta)=\frac92$ ✓ — correct radius, valid parametrization. …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.A straight rod of length 4 units slides such that its ends 'A' and 'B' always lie on the $x$ and $y$ axes respectively. Then the locus of the centroid of $\triangle OAB$ is ________ (A) $x^2+y^2=4$ (B) $x^2+y^2=3$ (C) $x^2+y^2=\dfrac{9}{16}$ (D) $x^2+y^2=\dfrac{16}{9}$

   ›Reveal solutionSolution

   This tests the locus of a centroid as a sliding rod's endpoints move along the axes; the constraint $a^2+b^2=16$ (constant rod length) becomes a circle in the centroid's coordinates.

   Concept and Intuition

   As the rod slides, $A=(a,0)$ and $B=(0,b)$ change, but the length $AB=4$ is fixed, so $a^2+b^2=16$ always. The centroid of $\triangle OAB$ (with $O$ the origin) is the average of the three vertices, and since two of them are on the axes, this average scales the constraint into a new circle.

   Step-by-Step Solution

   1. Let $A=(a,0)$ and $B=(0,b)$ be the points where the rod meets the axes. Since the rod has length 4: $a^2+b^2=16$.
   2. The centroid of $\triangle OAB$, with $O=(0,0)$, is $\left(\dfrac{0+a+0}{3},\dfrac{0+0+b}{3}\right)=\left(\dfrac{a}{3},\dfrac{b}{3}\right)$. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.The slope of the normal to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at $(x_1, y_1)$ is (A) $-\left(\dfrac{x_1+g}{y_1+f}\right)$ (B) $-\left(\dfrac{y_1+f}{x_1+g}\right)$ (C) $\dfrac{x_1+g}{y_1+f}$ (D) $\dfrac{y_1+f}{x_1+g}$

   ›Reveal solutionSolution

   The normal to a circle at any point always passes through the centre, so its slope is just the slope of the line joining the centre to that point: $\dfrac{y_1+f}{x_1+g}$.

   Concept and Intuition

   For a circle, the radius at a point is perpendicular to the tangent there, and the normal (being perpendicular to the tangent) is exactly the line through that point and the centre. So there is no need for calculus — the normal's slope is simply the slope of the radius.

   Step-by-Step Solution

   1. Write the circle in standard form: $x^2+y^2+2gx+2fy+c=0$ has centre $(-g,-f)$.
   2. The normal at $(x_1,y_1)$ is the line joining $(-g,-f)$ and $(x_1,y_1)$. …

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5. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If a circle of a constant radius 6 passes through origin $O$ and meets the coordinate axes at $A$ and $B$, then find the locus of the centroid of triangle $OAB$. (A) $x^2+y^2=4$ (B) $x^2+y^2=36$ (C) $x^2+y^2=16$ (D) $x^2+y^2=6$

   ›Reveal solutionSolution

   Since the circle passes through the origin and cuts the axes at $A,B$, the angle at $O$ is $90°$, making $AB$ a diameter; this fixes $a^2+b^2$ and the centroid's locus follows exactly as in Q1's pattern.

   Concept and Intuition

   Any angle inscribed in a semicircle is $90°$. Here $O$ lies on the circle and $OA,OB$ are along the two perpendicular axes, so $\angle AOB=90°$ is inscribed in the circle, meaning $AB$ must be a diameter.

   Step-by-Step Solution

   1. Circle has radius $6$, so its diameter is $12$.
   2. $A=(a,0)$, $B=(0,b)$ are both on the circle, and $O=(0,0)$ is also on the circle (given). Since $\angle AOB=90°$, $AB$ subtends a right angle at a point on the circle, so $AB$ must be a diameter: $AB=12$.
   3. By the distance formula, $AB^2=a^2+b^2=144$. …

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6. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x + y = 1$. If the intercepts made by the circle $x^2 + y^2 - x + 3y = 0$ on $L_1$ and $L_2$ are equal, then which of the following equations represent $L_1$ ____ (A) $x + y = 0$ & $x + 7y = 0$ (B) $x - y = 0$ & $x + 7y = 0$ (C) $x - 7y = 0$ & $x + y = 0$ (D) $x - 7y = 0$ & $x - y = 0$

   ›Reveal solutionSolution

   Equal intercepts on two lines from the same circle forces the centre to be equidistant

   from both lines; solving that distance-equality for the unknown slope through the

   origin gives $L_1: x-y=0$ or $x+7y=0$ — option (B).

   Concept and Intuition

   For a fixed circle of radius $r$, the length of the chord a line cuts off depends only on

   the perpendicular distance $d$ from the centre to that line: chord length

   $=2\sqrt{r^2-d^2}$. So "equal intercepts on $L_1$ and $L_2$" is exactly the same

   condition as "centre is equidistant from $L_1$ and $L_2$" — no need to compute the chord

   lengths themselves.

   Step-by-Step Solution

   1. Circle: $x^2+y^2-x+3y=0$. Centre $=\left(\dfrac12,-\dfrac32\right)$.
   2. Distance from centre to $L_2: x+y-1=0$:

   $d_2=\frac{\left|\tfrac12-\tfrac32-1\right|}{\sqrt2}=\frac{2}{\sqrt2}=\sqrt2$

   3. Let $L_1: y=mx$ i.e. $mx-y=0$ (a line through the origin). Distance from the centre to $L_1$:

   $d_1=\frac{\left|\tfrac{m}{2}+\tfrac32\right|}{\sqrt{m^2+1}}$

   4. Set $d_1=d_2=\sqrt2$ and square: $\left(\frac{m+3}{2}\right)^2=2(m^2+1) \Rightarrow (m+3)^2=8m^2+8$ …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ and its conjugate hyperbola, then the line $\dfrac{x}{2e_1} + \dfrac{y}{2e_2} = 1$ touches the circle having centre at the origin, then its radius is (A) $2$ (B) $e_1 + e_2$ (C) $e_1 e_2$ (D) $4$

   ›Reveal solutionSolution

   Using the standard identity $1/e_1^2+1/e_2^2=1$ for a hyperbola and its conjugate, the perpendicular distance from the origin to the given line collapses to a clean constant — that distance is the radius of the circle it touches.

   Concept and Intuition

   For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $e_1^2=1+\dfrac{b^2}{a^2}$, and for its conjugate $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$, $e_2^2=1+\dfrac{a^2}{b^2}$. So $\dfrac{1}{e_1^2}=\dfrac{a^2}{a^2+b^2}$ and $\dfrac1{e_2^2}=\dfrac{b^2}{a^2+b^2}$, and adding gives exactly $1$. A line always touches a circle centred at the origin when its distance from the origin equals the circle's radius.

   Step-by-Step Solution

   1. Identity: $\dfrac{1}{e_1^2}+\dfrac{1}{e_2^2}=1$.
   2. Line: $\dfrac{x}{2e_1}+\dfrac{y}{2e_2}-1=0$, i.e. $\dfrac{x}{2e_1}+\dfrac{y}{2e_2}=1$.
   3. Distance from origin $=\dfrac{|{-1}|}{\sqrt{\left(\frac{1}{2e_1}\right)^2+\left(\frac{1}{2e_2}\right)^2}}=\dfrac{1}{\sqrt{\frac{1}{4e_1^2}+\frac{1}{4e_2^2}}}=\dfrac{1}{\sqrt{\frac14\left(\frac1{e_1^2}+\frac1{e_2^2}\right)}}$. …

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8. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.Number of circles intersecting $x^2 + y^2 = 4$, $x^2 + y^2 - 2x - 3 = 0$ and $x^2 + y^2 - 2y - 3 = 0$ orthogonally is (A) $0$ (B) $1$ (C) $2$ (D) $\infty$

   ›Reveal solutionSolution

   The orthogonality conditions with three circles force a unique candidate circle, and it turns out to have negative radius$^2$ — so no real circle exists.

   Concept and Intuition

   Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ are orthogonal iff $2(g_1g_2+f_1f_2)=c_1+c_2$. Requiring orthogonality with three given circles gives three linear equations in the unknown circle's $(g,f,c)$ — generically an exactly-determined (unique) system.

   Step-by-Step Solution

   1. Let the sought circle be $x^2+y^2+2gx+2fy+c=0$.
   2. Orthogonal to $S_1$ ($g_1=0,f_1=0,c_1=-4$): $0=c-4\Rightarrow c=4$.
   3. Orthogonal to $S_2$ ($g_2=-1,f_2=0,c_2=-3$): $-2g=c-3=1\Rightarrow g=-\tfrac12$.
   4. Orthogonal to $S_3$ ($g_3=0,f_3=-1,c_3=-3$): $-2f=c-3=1\Rightarrow f=-\tfrac12$. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The straight line touching the circle $x^2 + y^2 - 2x - 3 = 0$ and remaining normal to the circle $x^2 + y^2 - 4y - 6 = 0$ is (A) $4x - 3y + 6 = 0$ (B) $y + 2 = 0$ (C) $4x + 3y - 6 = 0$ (D) $2x + 3 = 0$

   ›Reveal solutionSolution

   A line "normal to a circle" must pass through its centre; combine that with the tangency (distance = radius) condition on the other circle to test each option.

   Concept and Intuition

   The normal to a circle at any point on it always passes through the centre — that's the geometric definition of a normal line to a circle. So "normal to circle 2" simply means the line passes through circle 2's centre. Separately, "touching circle 1" means tangent to it, i.e. the perpendicular distance from circle 1's centre equals its radius.

   Step-by-Step Solution

   1. Circle 1: $x^2+y^2-2x-3=0\Rightarrow (x-1)^2+y^2=4$: centre $(1,0)$, radius $2$.
   2. Circle 2: $x^2+y^2-4y-6=0\Rightarrow x^2+(y-2)^2=10$: centre $(0,2)$.
   3. The line must pass through $(0,2)$ (normal to circle 2) and be tangent to circle 1 (distance from $(1,0)$ = 2).
   4. Test $4x-3y+6=0$: passes through $(0,2)$? $4(0)-3(2)+6=0$ ✓. Distance from $(1,0)$: $\dfrac{|4(1)-3(0)+6|}{\sqrt{16+9}}=\dfrac{10}{5}=2$ ✓ — matches the radius exactly. …

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10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.If $y = mx + c$ $(m > 0)$ is a common tangent to the parabola $y^2 = 12x$ and the circle $x^2 + y^2 = 36$, then $\dfrac{c^2}{m^2} =$ (A) $\dfrac{9}{3 - 2\sqrt{2}}$ (B) $36(3 + 2\sqrt{2})$ (C) $9(3 + 2\sqrt{2})$ (D) $\dfrac{36}{3 + 2\sqrt{2}}$

    ›Reveal solutionSolution

    Applying both tangency conditions (to the parabola and to the circle) and eliminating $m$ gives $c^2/m^2=36(3+2\sqrt2)$.

    Concept and Intuition

    A line $y=mx+c$ is tangent to $y^2=4ax$ iff $c=\dfrac{a}{m}$, and tangent to a circle $x^2+y^2=r^2$ iff the perpendicular distance from the center equals $r$: $\dfrac{|c|}{\sqrt{1+m^2}}=r \Rightarrow c^2=r^2(1+m^2)$. Being a common tangent means both conditions hold simultaneously for the same $m,c$.

    Step-by-Step Solution

    1. Parabola $y^2=12x \Rightarrow 4a=12 \Rightarrow a=3$. Tangency: $c=\dfrac{3}{m} \Rightarrow c^2=\dfrac{9}{m^2}$.
    2. Circle $x^2+y^2=36 \Rightarrow r=6$. Tangency: $c^2=36(1+m^2)$.
    3. Equate: $\dfrac{9}{m^2}=36(1+m^2) \Rightarrow 9=36m^2+36m^4 \Rightarrow 4m^4+4m^2-1=0$.
    4. Solve as quadratic in $m^2$: $m^2=\dfrac{-4\pm\sqrt{16+16}}{8}=\dfrac{-1\pm\sqrt2}{2}$. Since $m^2>0$, take $m^2=\dfrac{\sqrt2-1}{2}$.
    5. $m^4 = \left(\dfrac{\sqrt2-1}{2}\right)^2=\dfrac{3-2\sqrt2}{4}$. …

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11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.A(a, 0) is a fixed point and $\theta$ is a parameter such that $0 < \theta < 2\pi$. If P($a\cos\theta, a\sin\theta$) is a point on the circle $x^2+y^2=a^2$ and Q($b\sin\theta, -b\cos\theta$) is a point on the circle $x^2+y^2=b^2$, then the locus of the centroid of the triangle APQ is (A) a circle with centre at $\left(\dfrac{a}{3},0\right)$ and radius $\left(\dfrac{\sqrt{a^2+b^2}}{3}\right)$ (B) a circle with centre at $(a,0)$ and radius $\left(\dfrac{\sqrt{a^2+b^2}}{3}\right)$ (C) a parabola with focus at $\left(\dfrac{a}{3},0\right)$ (D) a parabola with focus at $(a,0)$

    ›Reveal solutionSolution

    Eliminating the parameter $\theta$ from the centroid's coordinates yields a circle centred at $(a/3,0)$ with radius $\sqrt{a^2+b^2}/3$ — option (A).

    Concept and Intuition

    When a locus is described parametrically via a single parameter $\theta$, the standard technique is to isolate the trigonometric parts on one side and use $\cos^2\theta+\sin^2\theta=1$ (often via squaring and adding two expressions) to eliminate $\theta$ entirely, revealing the underlying curve.

    Step-by-Step Solution

    1. Centroid of $A(a,0)$, $P(a\cos\theta,a\sin\theta)$, $Q(b\sin\theta,-b\cos\theta)$:

    $x=\frac{a+a\cos\theta+b\sin\theta}{3},\qquad y=\frac{a\sin\theta-b\cos\theta}{3}.$

    2. Rearrange: $3x-a = a\cos\theta+b\sin\theta$ and $3y=a\sin\theta-b\cos\theta$.
    3. Square and add:

    $(3x-a)^2+(3y)^2=(a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2.$

    4. Expand the right side: $a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2\sin^2\theta+a^2\sin^2\theta-2ab\sin\theta\cos\theta+b^2\cos^2\theta$. The cross terms cancel, leaving $a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)=a^2+b^2$. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.From a point P on the circle $x^2 + y^2 = 4$, two tangents are drawn to the circle $x^2 + y^2 - 6x - 6y + 14 = 0$. If A and B are the points of contact of those lines, then the locus of the centre of the circle passing through the points P, A and B is (A) $x^2 + y^2 - 3x - 3y + 4 = 0$ (B) $2x^2 + 2y^2 + 6x + 6y - 7 = 0$ (C) $x^2 + y^2 + 3x + 3y - 4 = 0$ (D) $2x^2 + 2y^2 - 6x - 6y + 7 = 0$

    ›Reveal solutionSolution

    The circle through $P$ and the two tangent points $A,B$ is exactly the circle with diameter $P C_2$ (where $C_2$ is the centre of the second circle), because the tangent-radius right angles at $A,B$ subtend that diameter; the locus of its centre (the midpoint of $P$ and $C_2$) is then found by substitution.

    Concept and Intuition

    Whenever tangents are drawn from an external point $P$ to a circle with centre $C$, touching it at $A$ and $B$, the radii $CA$ and $CB$ are perpendicular to the tangent lines $PA, PB$. So $\angle PAC=\angle PBC=90^\circ$. By the converse of the theorem "angle in a semicircle is a right angle", any point that sees segment $PC$ at $90^\circ$ lies on the circle having $PC$ as diameter. Hence $A$ and $B$ (and trivially $P$ itself) all lie on the circle with diameter $PC$ — this is the unique circle through $P,A,B$, with no further construction needed.

    Step-by-Step Solution

    1. Circle 1 (locus of $P$): $x^2+y^2=4$, centre $O=(0,0)$, radius $2$.
    2. Circle 2: $x^2+y^2-6x-6y+14=0 \Rightarrow$ centre $C_2=(3,3)$, radius$^2=3^2+3^2-14=4$, radius $2$.
    3. Let $P=(x_0,y_0)$ with $x_0^2+y_0^2=4$. The circle through $P,A,B$ has diameter $PC_2$, so its centre is the midpoint

    $\left(x,y\right)=\left(\frac{x_0+3}{2},\frac{y_0+3}{2}\right).$

    4. Solve for $x_0,y_0$: $x_0=2x-3,\; y_0=2y-3$. …

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