Q.If $\left|\dfrac{z-2}{z+2}\right|=\dfrac{\pi}{6}$, then the locus of $z$ is _____.

Showing the 12 most recent of 153 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The tangents drawn from the point $(-1,-1)$ to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$ touch the circle at the points A and B. If C is the centre of this circle, then the area of $\triangle ABC$ is (A) $3\sqrt{2}$ (B) $\dfrac{3}{\sqrt{2}}$ (C) $\dfrac{9}{2}$ (D) $9$

   ›Reveal solutionSolution

   Using the right triangle formed by the centre, external point, and a point of tangency, the chord-of-contact geometry gives triangle area $9/2$.

   Concept and Intuition

   For tangents from an external point $P$ to a circle of centre $C$ and radius $r$, touching at $A,B$: $CA\perp PA$ (radius $\perp$ tangent), so $\triangle CAP$ is right-angled at $A$. The chord of contact $AB$ is perpendicular to $CP$, and its midpoint $M$ divides $CP$ such that $CM=r^2/CP$ and $AM$ is the altitude of the right triangle to its hypotenuse.

   Step-by-Step Solution

   1. Circle $x^2+y^2-4x-4y-1=0$: centre $C=(2,2)$, radius$^2=4+4-(-1)=9\Rightarrow r=3$.
   2. Tangent length from $P(-1,-1)$: $PT^2 = (-1)^2+(-1)^2-4(-1)-4(-1)-1 = 1+1+4+4-1=9\Rightarrow PT=3$.
   3. $CP = \sqrt{(2-(-1))^2+(2-(-1))^2}=\sqrt{9+9}=3\sqrt2$.
   4. In right triangle $CAP$ (right angle at $A$), the altitude from $A$ onto hypotenuse $CP$ has foot $M$ (midpoint of chord of contact $AB$): $AM=\dfrac{CA\cdot AP}{CP}=\dfrac{3\cdot3}{3\sqrt2}=\dfrac{3}{\sqrt2}$, and $CM=\dfrac{CA^2}{CP}=\dfrac{9}{3\sqrt2}=\dfrac{3}{\sqrt2}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The power of a point $(1, 1)$ with respect to a circle S is 13 and the length of the tangent drawn from the point $(-1, 1)$ to the circle S is 1. If this circle S touches X–axis, then the radius of the circle S is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Use the algebraic definition of "power of a point" and "length of tangent" (both equal $S_1$, the circle's equation evaluated at the point) together with the tangency-to-X-axis condition to pin down the circle's coefficients.

   Concept and Intuition

   For a circle $S: x^2+y^2+2gx+2fy+c=0$, the power of any point $(x_1,y_1)$ is $S_1=x_1^2+y_1^2+2gx_1+2fy_1+c$, and the square of the tangent length from an external point is also $S_1$. A circle touching the X-axis has its center's distance to the X-axis equal to its radius, i.e. $|f|=r$, which combined with $r^2=g^2+f^2-c$ forces $c=g^2$.

   Step-by-Step Solution

   1. Power at $(1,1)$ is 13: $1+1+2g+2f+c=13 \Rightarrow 2g+2f+c=11$ … (I)
   2. Tangent length from $(-1,1)$ is 1, so $(\text{tangent})^2=1=$ power at $(-1,1)$: $1+1-2g+2f+c=1 \Rightarrow -2g+2f+c=-1$ … (II)
   3. Touches X-axis: center $=(-g,-f)$, distance to X-axis $=|f|=r$. Also $r^2=g^2+f^2-c$. So $f^2=g^2+f^2-c \Rightarrow c=g^2$.
   4. (I) $-$ (II): $4g = 12 \Rightarrow g=3$, so $c=g^2=9$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The line $x = \alpha$ touches the circle $x^2 + y^2 + 4x + 4y - 1 = 0$ at the point $P(h, k)$. If P lies in the fourth quadrant and $Q = (-2, 2)$, then $PQ =$ (A) $5$ (B) $4$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Identify the circle's center/radius, find the two vertical tangent lines, pick the one whose point of tangency lies in the fourth quadrant, then compute the distance to $Q$.

   Concept and Intuition

   A vertical tangent line to a circle touches it at the point having the same $y$-coordinate as the center, at $x=$ center's $x$-coordinate $\pm$ radius. Checking which of the two candidate tangency points falls in the fourth quadrant identifies $\alpha$ and $P$ uniquely.

   Step-by-Step Solution

   1. Rewrite the circle: $x^2+4x+y^2+4y=1 \Rightarrow (x+2)^2+(y+2)^2 = 1+4+4=9$. Center $=(-2,-2)$, radius $=3$.
   2. Vertical tangents occur at $x = -2+3=1$ and $x=-2-3=-5$, touching the circle at $(1,-2)$ and $(-5,-2)$ respectively (same $y$ as the center).
   3. Fourth quadrant requires $x>0, y<0$. Only $(1,-2)$ qualifies ($(-5,-2)$ has $x<0$, third quadrant). So $\alpha=1$ and $P=(1,-2)$. …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the equation of the Chord joining the points $P(\pi/6)$ and $P(\pi/3)$ on the circle $x^2 + y^2 + 6x - 4y + 9 = 0$ is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, then $a^2 + b^2 =$ (A) $4$ (B) $8$ (C) $6$ (D) $18$

   ›Reveal solutionSolution

   Parametrize the given circle, find the two named points explicitly, and derive the chord's equation directly from them — then read off the intercepts $a,b$.

   Concept and Intuition

   "$P(\theta)$" on a circle $(x-h)^2+(y-k)^2=r^2$ conventionally denotes the point $(h+r\cos\theta, k+r\sin\theta)$. Once the two actual points are computed, the chord joining them is just the line through two known points — no need for the general parametric chord formula, though it agrees.

   Step-by-Step Solution

   1. Circle: $x^2+y^2+6x-4y+9=0 \Rightarrow (x+3)^2+(y-2)^2 = 9+4-9=4$. Center $(-3,2)$, radius $2$.
   2. $P(\pi/6)$: $x=-3+2\cos\frac{\pi}{6}=-3+\sqrt3$, $y=2+2\sin\frac{\pi}{6}=2+1=3$. So $P_1=(-3+\sqrt3,\,3)$.
   3. $P(\pi/3)$: $x=-3+2\cos\frac{\pi}{3}=-3+1=-2$, $y=2+2\sin\frac{\pi}{3}=2+\sqrt3$. So $P_2=(-2,\,2+\sqrt3)$. …

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5. 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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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the chord $y = mx + 1$ of the circle $x^2 + y^2 = 1$ subtends an angle $45^\circ$ at the major segment of the circle, then the value of $m$ is (A) $2 \pm \sqrt{2}$ (B) $-2 \pm \sqrt{2}$ (C) $-1 \pm \sqrt{2}$ (D) $\pm 1$

   ›Reveal solutionSolution

   The chord subtends $45^\circ$ in the major segment, so by the inscribed-angle theorem the central angle is $90^\circ$; converting that into a perpendicular-distance condition on the line $y=mx+1$ gives $m=\pm1$. Answer: (D).

   Concept and Intuition

   The inscribed angle theorem says an angle subtended by a chord at any point on the major arc is half the central angle subtended by the same chord (on the minor-arc side). So a $45^\circ$ inscribed angle corresponds to a $90^\circ$ central angle. For a circle of radius $r$, if the central angle subtended by a chord is $2\alpha$, the perpendicular distance from the center to the chord is $r\cos\alpha$.

   Step-by-Step Solution

   1. Central angle $=2\times45^\circ=90^\circ$, so half-angle $\alpha=45^\circ$.
   2. For the unit circle ($r=1$), distance from center to chord $=r\cos\alpha=\cos45^\circ=\dfrac{\sqrt2}{2}$.
   3. Write the chord as $mx-y+1=0$. Distance from origin:

   $\frac{|0-0+1|}{\sqrt{m^2+1}}=\frac{1}{\sqrt{m^2+1}}$

   4. Set this equal to $\dfrac{\sqrt2}{2}$: …

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7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The tangent to the circle $C_1: x^2 + y^2 - 2x - 1 = 0$ at the point $(2,1)$ cuts off a chord of length 4 units from a circle $C_2$ whose centre is $(3,-2)$. The radius of the circle $C_2$ is (A) $\sqrt{6}$ (B) $2$ (C) $3$ (D) $2\sqrt{2}$

   ›Reveal solutionSolution

   Compute the tangent to $C_1$ at $(2,1)$, then use the perpendicular-distance/chord relation on $C_2$ to get $r=\sqrt6$. Answer: (A).

   Concept and Intuition

   The tangent to a circle at a known point on it can be written directly using the standard "replace $x^2\to xx_1$, $y^2\to yy_1$" tangent formula. Once we have that line, treating it as a chord of a second circle lets us use the classic right triangle relating radius, perpendicular distance from center to chord, and half the chord length: $r^2 = d^2 + (\text{half-chord})^2$.

   Step-by-Step Solution

   1. Circle $C_1: x^2+y^2-2x-1=0$ has $g=-1, f=0, c=-1$. Verify $(2,1)$ lies on it: $4+1-4-1=0$. ✓
   2. Tangent at $(x_1,y_1)=(2,1)$: $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$

   $2x+y-1(x+2)+0-1=0 \Rightarrow 2x+y-x-2-1=0 \Rightarrow x+y-3=0$

   3. This line cuts a chord of length $4$ from $C_2$, whose center is $(3,-2)$. Distance from center to line: $d=\frac{|3+(-2)-3|}{\sqrt{1^2+1^2}}=\frac{|-2|}{\sqrt2}=\sqrt2$ …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the radical axis of the circles $x^2 + y^2 + 2gx + 2fy + c = 0$ and $2x^2 + 2y^2 + 3x + 8y + 2c = 0$ touches the circle $x^2 + y^2 + 2x + 2y + 1 = 0$, then (A) $g = \dfrac{3}{2}$ or $f \neq 2$ (B) $g \neq \dfrac{3}{4}$ or $f = \dfrac{1}{2}$ (C) $g = \dfrac{1}{2}$ or $f = \dfrac{3}{4}$ (D) $g = \dfrac{3}{4}$ or $f = 2$

   ›Reveal solutionSolution

   Forming the radical axis (a line through the origin) and imposing tangency to the third circle reduces to $AB=0$, giving $g=3/4$ or $f=2$. Answer: (D).

   Concept and Intuition

   The radical axis of two circles $S_1=0$ and $S_2=0$ (each with unit coefficient of $x^2,y^2$) is simply $S_1-S_2=0$. Here the second circle must first be normalized (divided by 2) so both have coefficient $1$ for $x^2,y^2$. Tangency of a line to a circle is the standard condition: perpendicular distance from center equals radius.

   Step-by-Step Solution

   1. Normalize $2x^2+2y^2+3x+8y+2c=0$ by dividing by 2: $x^2+y^2+1.5x+4y+c=0$.
   2. Radical axis $=S_1-S_2$:

   $(2g-1.5)x+(2f-4)y+(c-c)=0 \Rightarrow (2g-1.5)x+(2f-4)y=0$

   This line passes through the origin.

   3. Circle $x^2+y^2+2x+2y+1=0$ has center $(-1,-1)$ and radius $\sqrt{1+1-1}=1$.

   4. Let $A=2g-1.5$, $B=2f-4$. Tangency requires:

   $\frac{|A(-1)+B(-1)|}{\sqrt{A^2+B^2}}=1 \Rightarrow \frac{|A+B|}{\sqrt{A^2+B^2}}=1$ …

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9. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The point of intersection of the lines $2x+3y-12=0$ and $3x-2y-5=0$ is the centre of a circle $S=0$. If $AB$ is a chord of $S=0$ and it is a diameter of the circle $x^2+y^2-10x+4y+13=0$, then the radius of the circle $S=0$ is (A) $4$ (B) $6$ (C) $16$ (D) $12$

   ›Reveal solutionSolution

   $AB$ is a diameter of the small circle, so its midpoint is that circle's centre; using the perpendicular-distance-from-centre relation for chord $AB$ of $S$ gives radius $6$.

   Concept and Intuition

   For a chord of a circle, the perpendicular distance $d$ from the circle's centre to the chord, together with the half-chord length $l$, satisfies $r^2=d^2+l^2$ (Pythagoras, since the perpendicular from the centre bisects the chord). Here the chord $AB$ of $S$ is also a diameter of a second, fully-known circle — so its midpoint (the second circle's centre) and half-length (that circle's radius) are both immediately available.

   Step-by-Step Solution

   1. Centre of $S$: solve $2x+3y-12=0$ and $3x-2y-5=0$. Multiply first by 2, second by 3: $4x+6y=24$, $9x-6y=15$; add: $13x=39\Rightarrow x=3$; then $y=2$. Centre of $S=(3,2)$.
   2. The circle $x^2+y^2-10x+4y+13=0$ has centre $(5,-2)$ and radius $\sqrt{25+4-13}=\sqrt{16}=4$.
   3. $AB$ is a diameter of this circle, so its midpoint is $(5,-2)$ and its half-length is $4$. …

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10. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.The centre of a circle $S=0$ is at $(2,5)$ and its radius is $r$. $S_1=0$ is a circle which lies in the second quadrant and touches the coordinate axes and intersects the circle $S=0$ at two points. If radius of circle $S_1=0$ is $2$, then the possible values of $r$ lie in the interval (A) $(2,8)$ (B) $(8,14)$ (C) $(3,7)$ (D) $(7,10)$

    ›Reveal solutionSolution

    $S_1$'s centre is fixed at $(-2,2)$ (touches both axes in Q2, radius 2); the two-intersection-point condition on the distance between centres gives $r\in(3,7)$.

    Concept and Intuition

    A circle touching both coordinate axes and lying in the second quadrant, with radius $a$, must have its centre at $(-a,a)$ (distance $a$ from each axis, in Q2). Two circles with centres a distance $d$ apart and radii $r_1,r_2$ intersect at exactly two points precisely when $|r_1-r_2| < d < r_1+r_2$ (strict inequalities — equality means tangency, and outside the range means no intersection).

    Step-by-Step Solution

    1. $S_1$ touches both axes in Q2 with radius $2$ $\Rightarrow$ centre $=(-2,2)$.
    2. Distance between centres of $S$ (at $(2,5)$) and $S_1$: $d=\sqrt{(2-(-2))^2+(5-2)^2}=\sqrt{16+9}=\sqrt{25}=5$.
    3. Two-intersection condition: $|r-2| < 5 < r+2$. …

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11. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.If the tangent drawn at $(\sqrt{2},1)$ on the circle $x^2+y^2=3$ is also a tangent to the two circles of equal radius $2\sqrt{3}$ with centres at $(0,\beta_1)$ and $(0,\beta_2)$, then $|\beta_1-\beta_2|=$ (A) $6$ (B) $12$ (C) $18$ (D) $5$

    ›Reveal solutionSolution

    Find the tangent line at the given point, then find the two centres on the y-axis equidistant ($=2\sqrt3$) from it; their separation is 12.

    Concept and Intuition

    The tangent to $x^2+y^2=r^2$ at $(x_1,y_1)$ is $xx_1+yy_1=r^2$ (the standard "replace $x^2\to xx_1$, $y^2\to yy_1$" rule). For this same line to be tangent to another circle, the perpendicular distance from that circle's centre to the line must equal its radius.

    Step-by-Step Solution

    1. Tangent to $x^2+y^2=3$ at $(\sqrt2,1)$: $\sqrt2\,x+1\cdot y=3$, i.e. $\sqrt2x+y-3=0$.
    2. Distance from $(0,\beta)$ to this line: $\dfrac{|\sqrt2(0)+\beta-3|}{\sqrt{2+1}}=\dfrac{|\beta-3|}{\sqrt3}$.
    3. Set equal to the given radius $2\sqrt3$: $\dfrac{|\beta-3|}{\sqrt3}=2\sqrt3\Rightarrow|\beta-3|=2\sqrt3\cdot\sqrt3=6$. …

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12. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $\theta$ is the angle between the tangents drawn to the circle $x^2+y^2-2x+4y+3=0$ from $(1,1)$ then, $|\tan\theta|=$ (A) $\dfrac{1}{\sqrt{2}}$ (B) $\dfrac{2\sqrt{14}}{5}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{\sqrt{14}}{5}$

    ›Reveal solutionSolution

    Use $\tan(\theta/2)=r/\sqrt{d^2-r^2}$ for the angle between the two tangents from an external point, then the double-angle formula gives $|\tan\theta|=\dfrac{2\sqrt{14}}{5}$.

    Concept and Intuition

    From an external point at distance $d$ from a circle's centre (radius $r$), the two tangent lines are symmetric about the line joining the point to the centre; each tangent makes a half-angle $\varphi$ with that line where $\sin\varphi=r/d$ (opposite side over hypotenuse in the right triangle formed by the centre, the point, and the point of tangency). The full angle between the tangents is $\theta=2\varphi$.

    Step-by-Step Solution

    1. Circle: $x^2+y^2-2x+4y+3=0\Rightarrow$ centre $(1,-2)$, radius$^2=1^2+2^2-3=2\Rightarrow r=\sqrt2$.
    2. Distance from $(1,1)$ to centre $(1,-2)$: $d=\sqrt{(1-1)^2+(1-(-2))^2}=\sqrt{0+9}=3$.
    3. Length of tangent from the point: $\sqrt{d^2-r^2}=\sqrt{9-2}=\sqrt7$.
    4. In the right triangle (centre–point–point of tangency), $\tan\varphi=\dfrac{r}{\sqrt{d^2-r^2}}=\dfrac{\sqrt2}{\sqrt7}$, where $\varphi=\theta/2$. …

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