Q.The differential equation $y\frac{dy}{dx}+x=c$ represents:
(A) Family of hyperbolas
(B) Family of parabolas
(C) Family of ellipses
(D) Family of circles

Showing the 12 most recent of 153 on this concept.

1. 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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2. 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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3. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If the angle between the circles $x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y+c=0$ is $\dfrac{\pi}{4}$, then $c=$ (A) $3$ (B) $-13$ (C) $-3$ or $13$ (D) $-31$ or $-3$

   ›Reveal solutionSolution

   Using the angle-between-circles formula and checking which root actually reproduces $\cos\theta=+1/\sqrt2$ (not its supplement) singles out $c=-13$.

   Concept and Intuition

   For two intersecting circles with radii $r_1, r_2$ and centre distance $d$, the angle $\theta$ between them at their point of intersection satisfies

   $\cos\theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$

   This is really just the law of cosines applied to the triangle formed by the two centres and a point of intersection (the radii there are along the normals to the tangents, so the angle between the tangents equals the angle between the radii). Because $\cos\theta$ can come out positive (acute angle) or negative (obtuse angle), it's essential to keep the correct sign rather than only solving the squared equation — squaring loses the sign information and can smuggle in a spurious extra root belonging to the supplementary angle $\pi-\theta$.

   Step-by-Step Solution

   1. Circle 1: $x^2+y^2+2x-4y+1=0 \Rightarrow$ centre $(-1,2)$, $r_1=\sqrt{1+4-1}=\sqrt4=2$.
   2. Circle 2: $x^2+y^2-4x-2y+c=0 \Rightarrow$ centre $(2,1)$, $r_2=\sqrt{4+1-c}=\sqrt{5-c}$.
   3. Distance between centres: $d=\sqrt{(2-(-1))^2+(1-2)^2}=\sqrt{9+1}=\sqrt{10}$, so $d^2=10$.
   4. Apply the formula with $\theta=\pi/4$, $\cos\theta = 1/\sqrt2$:

   $\frac{r_1^2+r_2^2-d^2}{2r_1r_2}=\frac{4+(5-c)-10}{2\cdot2\cdot\sqrt{5-c}}=\frac{-1-c}{4\sqrt{5-c}}=\frac{1}{\sqrt2}$

   5. Cross-multiply: $\sqrt2\,(-1-c) = 4\sqrt{5-c}$. For this to hold with a positive right side, we need $-1-c>0$, i.e. $c<-1$.
   6. Square both sides: $2(1+c)^2 = 16(5-c) \Rightarrow 2c^2+4c+2=80-16c \Rightarrow c^2+10c-39=0$.
   7. Solve: $c=\dfrac{-10\pm\sqrt{100+156}}{2}=\dfrac{-10\pm16}{2}=3\text{ or }-13$. …

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4. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If the circles $x^2+y^2-2x+4y+c=0$ and $x^2+y^2+2x-4y+c=0$ have four common tangents, then (A) $c<0$ (B) $-2<c<2$ (C) $0<c<5$ (D) $c>0$

   ›Reveal solutionSolution

   Both circles have the same radius $\sqrt{5-c}$ and their centres are $2\sqrt5$ apart; four common tangents exist only when the circles are entirely separate, i.e. centre-distance exceeds the sum of radii, giving $0<c<5$.

   Concept and Intuition

   Two circles have four common tangents (2 direct + 2 transverse) exactly when they lie completely outside each other, i.e. the distance between centres is greater than the sum of the radii.

   Step-by-Step Solution

   1. Circle 1: centre $(1,-2)$, $r_1^2=1+4-c=5-c$.
   2. Circle 2: centre $(-1,2)$, $r_2^2=1+4-c=5-c$ (same radius as circle 1).
   3. Distance between centres $=\sqrt{(1-(-1))^2+(-2-2)^2}=\sqrt{4+16}=2\sqrt5$.
   4. Condition for 4 common tangents: $2\sqrt5 > r_1+r_2=2\sqrt{5-c} \Rightarrow \sqrt5>\sqrt{5-c}\Rightarrow c>0$. …

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5. 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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6. 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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7. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Let $\theta$ be the angle between the circles $S \equiv x^2 + y^2 + 2x - 2y + c = 0$ and $S^1 \equiv x^2 + y^2 - 6x - 8y + 9 = 0$. If c is an integer and $\cos\theta = \frac{5}{16}$ then the radius of the circle $S = 0$ is (A) $2$ (B) $4$ (C) $3$ (D) $1$

   ›Reveal solutionSolution

   Solving $\cos\theta = \tfrac{5}{16}$ for the two circles gives radius $r_1 = 2$ for $S = 0$.

   Concept and Intuition

   For two circles with radii $r_1, r_2$ whose centres are a distance $d$ apart, the angle of intersection satisfies $\cos\theta = \dfrac{d^2 - r_1^2 - r_2^2}{2r_1 r_2}$. We know $S^1$ completely, so this becomes one equation in the unknown $r_1$.

   Step-by-Step Solution

   1. $S^1$: centre $(3,4)$, radius $r_2 = \sqrt{9+16-9} = 4$.
   2. $S$: centre $(-1,1)$, radius $r_1 = \sqrt{2-c}$, so $r_1^2 = 2-c$.
   3. Distance between centres: $d^2 = (3+1)^2 + (4-1)^2 = 16+9 = 25$.
   4. $\cos\theta = \dfrac{25 - (2-c) - 16}{2\cdot r_1 \cdot 4} = \dfrac{7+c}{8r_1} = \dfrac{5}{16}$.
   5. So $7+c = 2.5\,r_1$. Substitute $c = 2 - r_1^2$: $9 - r_1^2 = 2.5\,r_1$.
   6. $r_1^2 + 2.5r_1 - 9 = 0 \Rightarrow r_1 = \dfrac{-2.5 + 6.5}{2} = 2$ (taking the positive root). Then $c = 2 - 4 = -2$, an integer, as required.

   Common Mistakes …

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8. 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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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The set of all real values of c for which the equation $z\bar{z} + (4-3i)\bar{z} + (4+3i)z + c = 0$ represents a circle is (A) $[25, \infty)$ (B) $[-5, 5]$ (C) $(-\infty, -5] \cup [5, \infty)$ (D) $(-\infty, 25]$

   ›Reveal solutionSolution

   This tests the standard complex-number equation of a circle, $z\bar z + \bar a z + a\bar z + b=0$, whose radius² equals $|a|^2-b$; requiring a real, non-negative radius² pins down $c\le 25$.

   Concept and Intuition

   The general equation of a circle in the complex plane can be written as $z\bar z + \bar a z + a \bar z + b = 0$, where $b$ is real, and it represents a circle centred at $-a$ with radius $r=\sqrt{|a|^2 - b}$ (this comes directly from writing $z=x+iy$ and completing the square, or equivalently from $|z-(-a)|^2 = |a|^2-b$). For the equation to genuinely describe a circle in the real plane, the quantity under the square root, $|a|^2-b$, must be non-negative (a negative value would give no real locus at all, and a value of exactly zero gives a degenerate 'point circle').

   Step-by-Step Solution

   1. Compare the given equation $z\bar z + (4-3i)\bar z + (4+3i)z + c=0$ to the standard form $z\bar z + \bar a z + a\bar z + b=0$.
   2. Matching coefficients: the coefficient of $z$ is $a = 4+3i$; the coefficient of $\bar z$ should then be $\bar a = 4-3i$, which indeed matches the given equation. And $b = c$.
   3. Compute $|a|^2 = |4+3i|^2 = 4^2+3^2 = 16+9=25$.
   4. The radius² of the circle is $|a|^2 - b = 25 - c$.
   5. For the equation to represent a (real) circle, we require radius² to be non-negative: $25-c \ge 0 \Rightarrow c \le 25$.
   6. So the set of valid values of $c$ is $c \in (-\infty, 25]$. …

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10. 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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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Let C be the centre and A be one end of a diameter of the circle $x^2+y^2-2x-4y-20=0$. If P is a point on AC such that A divides CP in the ratio 2:3, then the locus of P is (A) $x^2+y^2-2x-4y-205=0$ (B) $2x^2+2y^2-4x-8y-405=0$ (C) $x^2+y^2-2x-4y-450=0$ (D) $4x^2+4y^2-8x-16y-605=0$

    ›Reveal solutionSolution

    As $A$ ranges over the circle, the section-formula relation shows $P$ always stays at a fixed larger distance $12.5$ from the same centre, so the locus is a concentric circle — giving $4x^2+4y^2-8x-16y-605=0$.

    Concept and Intuition

    Since $A$ is any point on the given circle, its distance from the centre $C$ is fixed (equal to the radius). The section-formula condition relating $P$ to $C$ and $A$ scales this fixed distance by a constant factor, so $P$ traces another circle centred at the same point $C$.

    Step-by-Step Solution

    1. Circle: $x^2+y^2-2x-4y-20=0\Rightarrow (x-1)^2+(y-2)^2=25$. So $C=(1,2)$, $r=5$.
    2. $A$ divides $CP$ in ratio $2:3$ (from $C$), so by the section formula $A=\dfrac{3C+2P}{5}\Rightarrow P=\dfrac{5A-3C}{2}$.
    3. Write $A=C+5u$ where $u$ is a unit vector (since $|CA|=r=5$).
    4. Then $P=\dfrac{5(C+5u)-3C}{2}=\dfrac{2C+25u}{2}=C+12.5u$.
    5. So $|P-C|=12.5$, i.e. locus is the circle $(x-1)^2+(y-2)^2=(12.5)^2=156.25$. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.If the circle $x^2 + y^2 + 2\alpha x + c = 0$ lies completely inside the circle $x^2 + y^2 + 2\beta x + c = 0$, then which of the following holds? (A) $\alpha\beta < 0$ (B) $c < 0$ (C) $c = 0$ (D) $\alpha\beta > 0$

    ›Reveal solutionSolution

    Both circles share the same constant term $c$, so they share the $y$-axis as their radical axis; for one circle to be strictly inside the other, the necessary relation between $\alpha$ and $\beta$ works out to $\alpha\beta>0$ — their centers must lie on the same side of the origin.

    Concept and Intuition

    Both circles $x^2+y^2+2\alpha x+c=0$ and $x^2+y^2+2\beta x + c=0$ have centers on the $x$-axis at $(-\alpha,0)$ and $(-\beta,0)$ and the same $c$. Subtracting the two equations kills the quadratic terms and leaves $2(\alpha-\beta)x=0$, i.e. the radical axis is the $y$-axis — independent of $\alpha,\beta$. For one circle to sit entirely inside the other (not crossing it), the distance between centers must be small enough compared to the difference of the radii.

    Step-by-Step Solution

    1. Centers: $C_1=(-\alpha,0)$, radius $r_1=\sqrt{\alpha^2-c}$; $C_2=(-\beta,0)$, radius $r_2=\sqrt{\beta^2-c}$.
    2. Distance between centers: $d=|\alpha-\beta|$.
    3. "Completely inside" (no intersection, $S_1$ strictly interior to $S_2$) requires $d < r_2-r_1$ (with $r_2>r_1$).
    4. Squaring: $(\alpha-\beta)^2 < r_2^2-2r_1r_2+r_1^2 = (\alpha^2+\beta^2-2c)-2r_1r_2$.
    5. Expand the left side and cancel $\alpha^2+\beta^2$ from both sides: $-2\alpha\beta < -2c-2r_1r_2 \Rightarrow \alpha\beta > c+r_1r_2$. …

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