Q.Show that the complex number $z$, satisfying the condition $\arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{4}$ lies on a circle.

Showing the 12 most recent of 45 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.In a triangle ABC, if $r - r_1 + r_2 + r_3 = 2\sqrt{2}R$, $r + r_1 - r_2 + r_3 = 0$ and $b = 2\sqrt{2}$, then $a + c =$ (A) $5$ (B) $6$ (C) $2 + \sqrt{2}$ (D) $4$

   ›Reveal solutionSolution

   The problem uses standard triangle identities for exradii and inradius to simplify the given equations, leading to a relation between the sides and the circumradius. Solving yields $a + c = 4$.

   The key here is to work with the well-known formulas for the inradius $r$ and exradii $r_1, r_2, r_3$ of a triangle. These are expressed in terms of the area $\Delta$ and the semiperimeter $s$:

   $r = \frac{\Delta}{s}, \quad r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c}$

   The given equations involve sums and differences of these radii. A powerful trick is to combine them using the identity $r_1 + r_2 + r_3 - r = 4R$, which relates the sum of all four radii to the circumradius $R$. But here we have specific signed combinations, so we’ll manipulate each equation separately.

   1. Write the first equation $r - r_1 + r_2 + r_3 = 2\sqrt{2}R$ Add and subtract $r$ cleverly: rewrite as $(r + r_2 + r_3) - r_1 = 2\sqrt{2}R$. Using the identity $r_1 + r_2 + r_3 - r = 4R$, we have $r_2 + r_3 - r = 4R - r_1$. Substitute into the left side: $(r + r_2 + r_3) - r_1 = (r + (4R - r_1 + r)) - r_1$? That’s messy. Instead, directly use the known result: $r - r_1 + r_2 + r_3 = (r_2 + r_3) + (r - r_1)$. A cleaner path: from $r_1 + r_2 + r_3 - r = 4R$, we get $r_2 + r_3 = 4R + r - r_1$. So $r - r_1 + r_2 + r_3 = r - r_1 + (4R + r - r_1) = 4R + 2r - 2r_1$. Thus the first equation becomes:

   $4R + 2r - 2r_1 = 2\sqrt{2}R$

   Divide by 2:

   $2R + r - r_1 = \sqrt{2}R \quad \Rightarrow \quad r - r_1 = (\sqrt{2} - 2)R$

   2. Now the second equation $r + r_1 - r_2 + r_3 = 0$ Again use $r_2 + r_3 = 4R + r - r_1$ from above. Then $-r_2 + r_3 = (r_3 - r_2)$. That’s not directly helpful. Instead, rewrite the left side as $(r + r_1 + r_3) - r_2 = 0$. From $r_1 + r_2 + r_3 - r = 4R$, we have $r_1 + r_3 = 4R + r - r_2$. So $r + r_1 + r_3 = r + (4R + r - r_2) = 4R + 2r - r_2$. Then the equation becomes:

   $4R + 2r - r_2 - r_2 = 0 \quad \Rightarrow \quad 4R + 2r - 2r_2 = 0$

   Divide by 2:

   $2R + r - r_2 = 0 \quad \Rightarrow \quad r - r_2 = -2R$

   3. Combine the results

      From step 1: $r - r_1 = (\sqrt{2} - 2)R$

      From step 2: $r - r_2 = -2R$

      Subtract the second from the first: $(r - r_1) - (r - r_2) = (\sqrt{2} - 2)R - (-2R)$

      $\Rightarrow -r_1 + r_2 = (\sqrt{2} - 2 + 2)R = \sqrt{2}R$

      So $r_2 - r_1 = \sqrt{2}R$

   4. Express $r_1$ and $r_2$ in terms of sides

      $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$.

      Their difference:

   $\frac{\Delta}{s-b} - \frac{\Delta}{s-a} = \sqrt{2}R$

   Factor $\Delta$:

   $\Delta \left( \frac{1}{s-b} - \frac{1}{s-a} \right) = \sqrt{2}R$

   Simplify the bracket:

   $\frac{(s-a) - (s-b)}{(s-b)(s-a)} = \frac{b-a}{(s-b)(s-a)}$

   So:

   $\Delta \cdot \frac{b-a}{(s-b)(s-a)} = \sqrt{2}R$

   5. Use known identities Recall $\Delta = \frac{abc}{4R}$ and $s = \frac{a+b+c}{2}$. Also, $s-a = \frac{b+c-a}{2}$, $s-b = \frac{a+c-b}{2}$. Substitute $\Delta$:

   $\frac{abc}{4R} \cdot \frac{b-a}{(s-b)(s-a)} = \sqrt{2}R$

   Multiply both sides by $4R$:

   $abc \cdot \frac{b-a}{(s-b)(s-a)} = 4\sqrt{2}R^2$

   6. Plug in $b = 2\sqrt{2}$ Then $b-a = 2\sqrt{2} - a$. Also $s-b = \frac{a + 2\sqrt{2} + c}{2} - 2\sqrt{2} = \frac{a + c - 2\sqrt{2}}{2}$. And $s-a = \frac{2\sqrt{2} + c - a}{2}$. So $(s-b)(s-a) = \frac{(a + c - 2\sqrt{2})(2\sqrt{2} + c - a)}{4}$. The numerator $abc = a \cdot 2\sqrt{2} \cdot c = 2\sqrt{2} ac$. The equation becomes:

   $2\sqrt{2} ac \cdot \frac{2\sqrt{2} - a}{\frac{(a + c - 2\sqrt{2})(2\sqrt{2} + c - a)}{4}} = 4\sqrt{2}R^2$

   Simplify the fraction: multiply numerator and denominator:

   $2\sqrt{2} ac \cdot (2\sqrt{2} - a) \cdot \frac{4}{(a + c - 2\sqrt{2})(2\sqrt{2} + c - a)} = 4\sqrt{2}R^2$

   Cancel $\sqrt{2}$ on both sides (divide by $\sqrt{2}$):

   $2 ac \cdot (2\sqrt{2} - a) \cdot \frac{4}{(a + c - 2\sqrt{2})(2\sqrt{2} + c - a)} = 4R^2$

   So:

   $\frac{8ac(2\sqrt{2} - a)}{(a + c - 2\sqrt{2})(2\sqrt{2} + c - a)} = 4R^2$

   7. Notice symmetry The denominator $(a + c - 2\sqrt{2})(2\sqrt{2} + c - a) = (c + (a - 2\sqrt{2}))(c + (2\sqrt{2} - a)) = c^2 - (a - 2\sqrt{2})^2$ This is $c^2 - (a^2 - 4\sqrt{2}a + 8) = c^2 - a^2 + 4\sqrt{2}a - 8$. The numerator $8ac(2\sqrt{2} - a)$ is $8ac(2\sqrt{2} - a)$. This looks messy. Perhaps there’s a simpler route.

   Tip

   Instead of expanding, use the relation $r - r_2 = -2R$ directly. Since $r = \frac{\Delta}{s}$ and $r_2 = \frac{\Delta}{s-b}$, we have $\frac{\Delta}{s} - \frac{\Delta}{s-b} = -2R$. Factor $\Delta$: $\Delta \left( \frac{1}{s} - \frac{1}{s-b} \right) = -2R$. Simplify: $\frac{(s-b) - s}{s(s-b)} = \frac{-b}{s(s-b)}$, so $-\Delta \cdot \frac{b}{s(s-b)} = -2R$, giving $\frac{\Delta b}{s(s-b)} = 2R$. But $\Delta = \frac{abc}{4R}$, so $\frac{abc}{4R} \cdot \frac{b}{s(s-b)} = 2R$ → $\frac{ab^2 c}{4R s(s-b)} = 2R$ → $ab^2 c = 8R^2 s(s-b)$. With $b=2\sqrt{2}$, $b^2=8$, so $8ac = 8R^2 s(s-b)$ → $ac = R^2 s(s-b)$.

   8. Use the first equation similarly

      From $r - r_1 = (\sqrt{2} - 2)R$, we get $\frac{\Delta}{s} - \frac{\Delta}{s-a} = (\sqrt{2} - 2)R$.

      Simplify: $\Delta \cdot \frac{-a}{s(s-a)} = (\sqrt{2} - 2)R$ → $-\frac{\Delta a}{s(s-a)} = (\sqrt{2} - 2)R$.

      Substitute $\Delta = \frac{abc}{4R}$: $-\frac{abc \cdot a}{4R s(s-a)} = (\sqrt{2} - 2)R$ → $-\frac{a^2 bc}{4R s(s-a)} = (\sqrt{2} - 2)R$ → $a^2 bc = -4(\sqrt{2} - 2)R^2 s(s-a) = 4(2 - \sqrt{2})R^2 s(s-a)$.

      With $b=2\sqrt{2}$, $bc = 2\sqrt{2}c$, so $a^2 \cdot 2\sqrt{2}c = 4(2-\sqrt{2})R^2 s(s-a)$ → $2\sqrt{2} a^2 c = 4(2-\sqrt{2})R^2 s(s-a)$.

   9. Divide the two results

      From step 7: $ac = R^2 s(s-b)$

      From step 8: $2\sqrt{2} a^2 c = 4(2-\sqrt{2})R^2 s(s-a)$

      Divide the second by the first:

   $\frac{2\sqrt{2} a^2 c}{ac} = \frac{4(2-\sqrt{2})R^2 s(s-a)}{R^2 s(s-b)}$

   $\Rightarrow 2\sqrt{2} a = \frac{4(2-\sqrt{2})(s-a)}{s-b}$

   So $a = \frac{2(2-\sqrt{2})(s-a)}{\sqrt{2}(s-b)} = \frac{2(2-\sqrt{2})}{\sqrt{2}} \cdot \frac{s-a}{s-b} = \sqrt{2}(2-\sqrt{2}) \cdot \frac{s-a}{s-b}$

   Since $\sqrt{2}(2-\sqrt{2}) = 2\sqrt{2} - 2$, we have:

   $a = (2\sqrt{2} - 2) \frac{s-a}{s-b}$

   10. Solve for $a$ and $c$ Recall $s = \frac{a+b+c}{2} = \frac{a + 2\sqrt{2} + c}{2}$. Then $s-a = \frac{a + 2\sqrt{2} + c}{2} - a = \frac{-a + 2\sqrt{2} + c}{2} = \frac{c - a + 2\sqrt{2}}{2}$ And $s-b = \frac{a + 2\sqrt{2} + c}{2} - 2\sqrt{2} = \frac{a + c - 2\sqrt{2}}{2}$ Substitute into the equation for $a$:

   $a = (2\sqrt{2} - 2) \cdot \frac{\frac{c - a + 2\sqrt{2}}{2}}{\frac{a + c - 2\sqrt{2}}{2}} = (2\sqrt{2} - 2) \cdot \frac{c - a + 2\sqrt{2}}{a + c - 2\sqrt{2}}$

   Cross-multiply:  
   

   $a(a + c - 2\sqrt{2}) = (2\sqrt{2} - 2)(c - a + 2\sqrt{2})$

   Expand left: $a^2 + ac - 2\sqrt{2}a$  
   Right: $(2\sqrt{2} - 2)c - (2\sqrt{2} - 2)a + (2\sqrt{2} - 2) \cdot 2\sqrt{2}$ …
   

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Let S, S' be the foci and B be one end of the minor axis of an ellipse. If $\angle SBS' = 120^\circ$ and the area of triangle SBS' is $\sqrt{3}$ then the length of its latus rectum is (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{\sqrt{3}}$ (C) $1$ (D) $\sqrt{3}$

   ›Reveal solutionSolution

   The key is to use the geometry of the ellipse: the foci and minor-axis vertex form an isosceles triangle. Using the given angle and area, we find $b$ and $e$, then the latus rectum. The length is $\boxed{1}$.

   The problem gives us an ellipse with foci $S$ and $S'$, and $B$ as one end of the minor axis. The triangle $SBS'$ has $\angle SBS' = 120^\circ$ and area $\sqrt{3}$. We need the length of the latus rectum.

   Let’s recall the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, foci at $(\pm ae, 0)$, and ends of the minor axis at $(0, \pm b)$. Here $e = \sqrt{1 - \frac{b^2}{a^2}}$, and the distance from centre to each focus is $ae$.

   The triangle $SBS'$ has vertices at $S(-ae, 0)$, $S'(ae, 0)$, and $B(0, b)$. Notice $SB = S'B$ because $B$ is equidistant from both foci (by symmetry of the ellipse). So triangle $SBS'$ is isosceles with vertex at $B$.

   1. Find the side lengths in terms of $a$, $b$, $e$. The base $SS' = 2ae$. The equal sides: $SB = \sqrt{(ae)^2 + b^2}$. But recall the ellipse property: $b^2 = a^2(1 - e^2)$. So

   $SB = \sqrt{a^2 e^2 + a^2(1 - e^2)} = \sqrt{a^2} = a.$

   So each equal side is simply $a$. That’s neat — the distance from a focus to an end of the minor axis equals the semi-major axis.

   2. Use the given angle $\angle SBS' = 120^\circ$. In triangle $SBS'$, sides: $SB = S'B = a$, base $SS' = 2ae$. Apply the cosine rule at vertex $B$:

   $(SS')^2 = SB^2 + S'B^2 - 2(SB)(S'B)\cos 120^\circ.$

   $\cos 120^\circ = -\frac{1}{2}$, so

   $(2ae)^2 = a^2 + a^2 - 2a^2\left(-\frac{1}{2}\right) = 2a^2 + a^2 = 3a^2.$

   Thus $4a^2 e^2 = 3a^2$, giving $e^2 = \frac{3}{4}$, so $e = \frac{\sqrt{3}}{2}$ (positive).

   3. Use the area condition. Area of triangle $SBS'$ is $\frac{1}{2} \times \text{base} \times \text{height}$. Base $SS' = 2ae = 2a \cdot \frac{\sqrt{3}}{2} = a\sqrt{3}$. Height from $B$ to the x-axis is $b$ (since $B$ is at $(0,b)$ and the base lies on the x-axis). So area = $\frac{1}{2} \times a\sqrt{3} \times b = \frac{\sqrt{3}}{2} a b$. Given area = $\sqrt{3}$, we have $$$$

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.In triangle ABC, $(r_1 - r)\cos\left(\dfrac{B-C}{2}\right)=$ (A) $(r_1 + r)\sin\dfrac{A}{2}$ (B) $(r_2 + r_3)\sin\dfrac{A}{2}$ (C) $(r_1 + r)\sin\dfrac{(B-C)}{2}$ (D) $(r_2 + r_3)\sin\dfrac{(B-C)}{2}$

   ›Reveal solutionSolution

   With the half-angle forms of $r$ and $r_1$, both $(r_1-r)\cos\frac{B-C}{2}$ and $(r_1+r)\sin\frac A2$ reduce to $4R\sin^2\frac A2\cos\frac{B-C}{2}$, so they are equal — option (A).

   Half-angle forms of the radii.

   $r=4R\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2,\qquad r_1=4R\sin\tfrac A2\cos\tfrac B2\cos\tfrac C2.$

   Difference.

   $r_1-r=4R\sin\tfrac A2\left(\cos\tfrac B2\cos\tfrac C2-\sin\tfrac B2\sin\tfrac C2\right)=4R\sin\tfrac A2\cos\tfrac{B+C}{2}.$

   Since $\tfrac{B+C}{2}=\tfrac\pi2-\tfrac A2$, we have $\cos\tfrac{B+C}{2}=\sin\tfrac A2$, hence

   $r_1-r=4R\sin^2\tfrac A2.$

   Sum. …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $z = x + iy$ and the point P denotes $z$ in the Argand plane, then the locus of P satisfying the condition $\operatorname{Im}\left(\dfrac{z-3i}{z+2}\right) = 1,\, z \neq -2$ is (A) $7x - 2y + 10 = 0$ (B) $x^2 + xy + y^2 + 7x - 2y + 10 = 0$ (C) $x^2 - xy + y^2 + 7x - 2y + 10 = 0$ (D) $x^2 + y^2 + 7x - 2y + 10 = 0$

   ›Reveal solutionSolution

   The condition $\operatorname{Im}\left(\frac{z-3i}{z+2}\right)=1$ forces the imaginary part of a complex fraction to equal 1. By writing $z=x+iy$, simplifying the fraction, and equating the imaginary part to 1, we obtain a linear equation — the locus is a straight line, not a circle or conic. The final equation is $7x - 2y + 10 = 0$.

   The key insight: when a problem says "Im(something) = constant", you are isolating only the imaginary part of a complex expression. That often yields a linear relation between $x$ and $y$, because the imaginary part of a rational function simplifies to a ratio of two linear expressions. Setting that ratio equal to a constant gives a linear equation — provided the denominator doesn't vanish.

   Let’s work it through.

   1. Write the fraction in terms of $x$ and $y$. Given $z = x + iy$, we have

   $z - 3i = x + i(y - 3), \quad z + 2 = (x + 2) + iy.$

   So

   $\frac{z-3i}{z+2} = \frac{x + i(y-3)}{(x+2) + iy}.$

   2. Rationalise to separate real and imaginary parts. Multiply numerator and denominator by the conjugate of the denominator:

   $\frac{x + i(y-3)}{(x+2) + iy} \times \frac{(x+2) - iy}{(x+2) - iy}.$

   The denominator becomes

   $(x+2)^2 + y^2.$

   The numerator expands as

   $[x + i(y-3)]\,[(x+2) - iy] = x(x+2) - ixy + i(y-3)(x+2) - i^2 y(y-3).$

   Since $i^2 = -1$, the last term becomes $+ y(y-3)$. So the numerator is

   $x(x+2) + y(y-3) + i\big[ -xy + (y-3)(x+2) \big].$

   3. Simplify the imaginary part. The coefficient of $i$ is

   $-xy + (y-3)(x+2) = -xy + (yx + 2y - 3x - 6).$

   The $-xy$ and $+yx$ cancel. We are left with

   $2y - 3x - 6.$

   So the imaginary part of the fraction is

   $\operatorname{Im}\left(\frac{z-3i}{z+2}\right) = \frac{2y - 3x - 6}{(x+2)^2 + y^2}.$

   4. Set this equal to 1. The given condition is

   $\frac{2y - 3x - 6}{(x+2)^2 + y^2} = 1.$

   Multiply through (the denominator is never zero for $z \neq -2$, so it's safe):

   $2y - 3x - 6 = (x+2)^2 + y^2.$

   5. Expand and simplify.

   $2y - 3x - 6 = x^2 + 4x + 4 + y^2.$

   Bring all terms to one side:

   $0 = x^2 + 4x + 4 + y^2 - 2y + 3x + 6.$

   Combine like terms:

   $x^2 + y^2 + (4x + 3x) + (-2y) + (4 + 6) = x^2 + y^2 + 7x - 2y + 10.$

   So we have

   $x^2 + y^2 + 7x - 2y + 10 = 0.$

   That looks like option (D). But wait — is this the final locus? Let’s check if it’s actually a circle. Complete the square:

   $(x^2 + 7x) + (y^2 - 2y) = -10$

   $\left(x + \frac{7}{2}\right)^2 - \frac{49}{4} + (y - 1)^2 - 1 = -10$ …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Area of the quadrilateral formed by the common tangents drawn to the circle $x^2 + y^2 = 16$ and the ellipse $7x^2 + 25y^2 = 175$ is (A) $64$ (B) $32$ (C) $5\sqrt{2}$ (D) $16\sqrt{2}$

   ›Reveal solutionSolution

   The four common tangents to the circle and ellipse form a square whose diagonals lie along the coordinate axes. Finding the tangent lines and their intersection points gives an area of $64$ — option (A).

   The problem asks for the area of the quadrilateral formed by the common tangents to a circle and an ellipse, both centred at the origin. Because both curves are centred at the origin and symmetric about both axes, the common tangents come in symmetric pairs, and the quadrilateral they enclose is a square rotated $45^\circ$ relative to the axes.

   1. Rewrite the equations in standard form. The circle is $x^2 + y^2 = 16$, so its centre is $(0,0)$ and radius $r = 4$. The ellipse is $7x^2 + 25y^2 = 175$. Divide through by $175$:

   $\frac{x^2}{25} + \frac{y^2}{7} = 1$

   So $a^2 = 25$, $b^2 = 7$, with $a > b$, also centred at the origin.

   2. Set up the tangency condition for the circle. Let a tangent line be $y = mx + c$. For it to be tangent to the circle $x^2 + y^2 = 16$, the perpendicular distance from the centre to the line must equal the radius:

   $\frac{|c|}{\sqrt{1+m^2}} = 4 \quad \Rightarrow \quad c^2 = 16(1+m^2)$

   3. Apply the tangency condition for the ellipse. For the ellipse $\frac{x^2}{25} + \frac{y^2}{7} = 1$, the condition that $y = mx + c$ is tangent is:

   $c^2 = a^2 m^2 + b^2 = 25m^2 + 7$

   4. Equate the two expressions for $c^2$.

   $16 + 16m^2 = 25m^2 + 7$

   $16 - 7 = 25m^2 - 16m^2$

   $9 = 9m^2 \quad \Rightarrow \quad m^2 = 1 \quad \Rightarrow \quad m = \pm 1$

   5. Find the corresponding $c$ values. Using $c^2 = 16(1+m^2) = 16(2) = 32$, so $c = \pm 4\sqrt{2}$. Thus the four common tangents are:

   $y = x + 4\sqrt{2}, \quad y = x - 4\sqrt{2}, \quad y = -x + 4\sqrt{2}, \quad y = -x - 4\sqrt{2}$

   6. Find the vertices of the quadrilateral they enclose. Solving the four lines pairwise for their intersections:
      • $y=x+4\sqrt2$ and $y=-x+4\sqrt2$ intersect at $(0, 4\sqrt{2})$.
      • $y=x+4\sqrt2$ and $y=-x-4\sqrt2$ intersect at $(-4\sqrt{2}, 0)$.
      • $y=x-4\sqrt2$ and $y=-x+4\sqrt2$ intersect at $(4\sqrt{2}, 0)$. …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Let the angles A, B, C of a triangle ABC be in arithmetic progression. If the exradii $r_1, r_2, r_3$ of triangle ABC satisfy the condition $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$, then $b =$ (A) $\dfrac{2a}{\sqrt{3}}$ (B) $\sqrt{2}a$ (C) $\sqrt{3}a$ (D) $a$

   ›Reveal solutionSolution

   The angles are in arithmetic progression, so the middle angle is 60°. Using exradii formulas and the given relation, we derive that side $b$ equals $\sqrt{3}a$. The correct option is (C).

   ---

   We are told that angles $A, B, C$ of a triangle are in arithmetic progression.

   That means $A + C = 2B$. Since $A + B + C = \pi$, we get

   $2B + B = \pi \quad\Rightarrow\quad 3B = \pi \quad\Rightarrow\quad B = \frac{\pi}{3}.$

   So angle $B$ is fixed at $60^\circ$. This is the key geometric fact.

   Now we are given a relation among the exradii:

   $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1.$

   We need to find which side relation follows.

   ---

   Concept and intuition:

   Exradii are related to the sides and area. The standard formulas are

   $r_1 = \frac{\Delta}{s-a},\quad r_2 = \frac{\Delta}{s-b},\quad r_3 = \frac{\Delta}{s-c},$

   where $\Delta$ is the area and $s$ the semiperimeter.

   The given equation is symmetric in $r_1, r_2, r_3$ except that $r_3$ appears squared. That suggests we can substitute these expressions and simplify using the fact that $B = 60^\circ$ to get a relation between sides.

   ---

   Step-by-step solution:

   1. Write the exradii in terms of sides and area.

   $r_1 = \frac{\Delta}{s-a},\quad r_2 = \frac{\Delta}{s-b},\quad r_3 = \frac{\Delta}{s-c}.$

   2. Substitute into the given equation $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$:

   $\frac{\Delta^2}{(s-c)^2} = \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}.$

   Cancel $\Delta^2$ (nonzero for a nondegenerate triangle).

   3. Multiply through by $(s-a)(s-b)(s-c)$ to clear denominators:

   $(s-a)(s-b) = (s-c)^2 + (s-a)(s-c) + (s-b)(s-c).$

   4. Expand and simplify. Left: $(s-a)(s-b) = s^2 - s(a+b) + ab$. Right:

   $(s-c)^2 = s^2 - 2sc + c^2,$

   $(s-a)(s-c) = s^2 - s(a+c) + ac,$

   $(s-b)(s-c) = s^2 - s(b+c) + bc.$

   Summing the three terms on the right:

   $3s^2 - s[(2c) + (a+c) + (b+c)] + (c^2 + ac + bc).$

   The coefficient of $s$: $2c + a + c + b + c = a + b + 4c$.

   So right side = $3s^2 - s(a+b+4c) + (c^2 + ac + bc)$.

   5. Equate left and right:

   $s^2 - s(a+b) + ab = 3s^2 - s(a+b+4c) + (c^2 + ac + bc).$

   Bring all terms to one side:

   $0 = 2s^2 - s(4c) + (c^2 + ac + bc - ab).$

   So

   $2s^2 - 4c s + (c^2 + ac + bc - ab) = 0.$

   6. Recall $s = \frac{a+b+c}{2}$. Substitute:

   $2\left(\frac{a+b+c}{2}\right)^2 - 4c\left(\frac{a+b+c}{2}\right) + (c^2 + ac + bc - ab) = 0.$

   Simplify:

   $\frac{(a+b+c)^2}{2} - 2c(a+b+c) + (c^2 + ac + bc - ab) = 0.$

   Multiply by 2:

   $(a+b+c)^2 - 4c(a+b+c) + 2(c^2 + ac + bc - ab) = 0.$

   7. Expand $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

      Then $-4c(a+b+c) = -4ac - 4bc - 4c^2$.

      And $2(c^2 + ac + bc - ab) = 2c^2 + 2ac + 2bc - 2ab$.

      Summing all:

   $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 4ac - 4bc - 4c^2 + 2c^2 + 2ac + 2bc - 2ab = 0.$

   Combine like terms: …

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7. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the angle between the circles $x^2 + y^2 - 2x + ky + 1 = 0$ and $x^2 + y^2 - kr - 2y + 1 = 0$ is $\cos^{-1}\left(\dfrac{1}{4}\right)$ and $k < 0$ then the point which lies on the radical axis of the given circles is (A) $(1, -3)$ (B) $(-1, 3)$ (C) $(-1, -3)$ (D) $(1, 3)$

   ›Reveal solutionSolution

   The key idea is to use the formula for the angle between two circles to solve for $k$, then find the radical axis equation, and finally test which point satisfies it. The correct point is $(1, -3)$.

   ---

   We start with the two circles:

   $C_1: x^2 + y^2 - 2x + ky + 1 = 0$

   $C_2: x^2 + y^2 - kx - 2y + 1 = 0$

   (Note: The problem originally had “$-kr$” which is a typo; it should be $-kx$.)

   ---

   Concept and Intuition

   The angle between two circles is defined as the angle between their tangents at a point of intersection. It can be found using their radii and the distance between their centers:

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

   where $d$ is the distance between centers, and $r_1, r_2$ are the radii.

   We are given $\theta = \cos^{-1}(1/4)$, so $\cos\theta = 1/4$.

   The radical axis of two circles is the line of points having equal power with respect to both circles. It is obtained by subtracting the equations.

   We will:

   1. Find centers and radii.
   2. Use the angle condition to solve for $k$ (with $k<0$).
   3. Write the radical axis.
   4. Test which given point lies on it.

   ---

   Step-by-step solution

   1. Find centers and radii of the circles.

   For $C_1$:

   Complete squares:

   $x^2 - 2x + y^2 + ky + 1 = 0$

   $(x-1)^2 - 1 + \left(y + \frac{k}{2}\right)^2 - \frac{k^2}{4} + 1 = 0$

   $(x-1)^2 + \left(y + \frac{k}{2}\right)^2 = \frac{k^2}{4}$

   So center $C_1 = (1, -k/2)$, radius $r_1 = |k|/2$.

   For $C_2$:

   $x^2 - kx + y^2 - 2y + 1 = 0$

   $\left(x - \frac{k}{2}\right)^2 - \frac{k^2}{4} + (y-1)^2 - 1 + 1 = 0$

   $\left(x - \frac{k}{2}\right)^2 + (y-1)^2 = \frac{k^2}{4}$

   So center $C_2 = (k/2, 1)$, radius $r_2 = |k|/2$.

   Both radii are equal: $r_1 = r_2 = |k|/2$.

   2. Distance between centers.

   $d = \sqrt{\left(1 - \frac{k}{2}\right)^2 + \left(-\frac{k}{2} - 1\right)^2}$

   Simplify:

   $d^2 = \left(1 - \frac{k}{2}\right)^2 + \left(-\frac{k}{2} - 1\right)^2$

   $= \left(1 - \frac{k}{2}\right)^2 + \left(1 + \frac{k}{2}\right)^2$

   $= 2\left(1 + \frac{k^2}{4}\right) = 2 + \frac{k^2}{2}$

   3. Use the angle condition.

   $\cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2} = \frac{1}{4}$

   Substitute:

   $r_1^2 = r_2^2 = \frac{k^2}{4}, \quad 2 r_1 r_2 = 2 \cdot \frac{k^2}{4} = \frac{k^2}{2}$

   So:

   $\frac{\left(2 + \frac{k^2}{2}\right) - \frac{k^2}{4} - \frac{k^2}{4}}{\frac{k^2}{2}} = \frac{1}{4}$

   Numerator simplifies:

   $2 + \frac{k^2}{2} - \frac{k^2}{2} = 2$

   Thus:

   $\frac{2}{k^2/2} = \frac{4}{k^2} = \frac{1}{4}$

   So:

   $k^2 = 16 \quad \Rightarrow \quad k = \pm 4$

   Given $k < 0$, we take $k = -4$.

   4. Find the radical axis.

   Subtract the equations of the circles (with $k = -4$):

   $C_1: x^2 + y^2 - 2x - 4y + 1 = 0$

   $C_2: x^2 + y^2 + 4x - 2y + 1 = 0$

   Subtract $C_2$ from $C_1$:

   $(-2x - 4y) - (4x - 2y) = 0$

   $-2x - 4y - 4x + 2y = 0$

   $-6x - 2y = 0 \quad \Rightarrow \quad 3x + y = 0$

   So the radical axis is the line $y = -3x$.

   5. Test the given points.

   • (A) $(1, -3)$: $-3 = -3(1)$ → lies on line.
   • (B) $(-1, 3)$: $3 = -3(-1) = 3$ → lies on line.
   • (C) $(-1, -3)$: $-3 = -3(-1) = 3$ → no.
   • (D) $(1, 3)$: $3 = -3(1) = -3$ → no. …

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.In a triangle ABC, if $r_1 = 4$, $r_2 = 8$ and $r_3 = 24$, then $a : b : c =$ (A) $4 : 7 : 9$ (B) $2 : 3 : 5$ (C) $1 : 2 : 6$ (D) $6 : 2 : 1$

   ›Reveal solutionSolution

   Using $s-a=\Delta/r_1$ etc., the sides come out proportional to $4:7:9$, option (A).

   For exradii, $r_1=\dfrac{\Delta}{s-a},\ r_2=\dfrac{\Delta}{s-b},\ r_3=\dfrac{\Delta}{s-c}$, so

   $s-a=\frac{\Delta}{4},\qquad s-b=\frac{\Delta}{8},\qquad s-c=\frac{\Delta}{24}.$

   Adding, $(s-a)+(s-b)+(s-c)=3s-2s=s$, hence

   $s=\Delta\left(\frac{6+3+1}{24}\right)=\frac{5\Delta}{12}.$

   Then each side is $s$ minus the corresponding difference: …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.In a triangle ABC, $(r_2 + r_3) \csc^2 \left(\dfrac{A}{2}\right) =$ (A) $4R \cot \left(\dfrac{A}{2}\right)$ (B) $2R \cot^2 \left(\dfrac{A}{2}\right)$ (C) $\dfrac{4R}{\tan^2 \left(\dfrac{A}{2}\right)}$ (D) $\dfrac{2R}{\tan \left(\dfrac{A}{2}\right)}$

   ›Reveal solutionSolution

   $(r_2+r_3)\csc^2\!\tfrac{A}{2}=4R\cot^2\!\tfrac{A}{2}=\dfrac{4R}{\tan^2(A/2)}$, option (C).

   Use the angle forms $r_2=4R\sin\tfrac{B}{2}\cos\tfrac{C}{2}\cos\tfrac{A}{2}$ and $r_3=4R\sin\tfrac{C}{2}\cos\tfrac{B}{2}\cos\tfrac{A}{2}$. Then

   $r_2+r_3 = 4R\cos\tfrac{A}{2}\Big(\sin\tfrac{B}{2}\cos\tfrac{C}{2}+\cos\tfrac{B}{2}\sin\tfrac{C}{2}\Big)=4R\cos\tfrac{A}{2}\,\sin\tfrac{B+C}{2}.$

   Since $\tfrac{B+C}{2}=\tfrac{\pi-A}{2}$, we have $\sin\tfrac{B+C}{2}=\cos\tfrac{A}{2}$, so

   $r_2+r_3 = 4R\cos^2\tfrac{A}{2}.$

   Multiplying by $\csc^2\tfrac{A}{2}$: …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.If the percentage error in the radius of a circle is 3, then the percentage error in its area is (A) 6 (B) $\dfrac{3}{2}$ (C) 2 (D) 4

    ›Reveal solutionSolution

    The percentage error in area is twice the percentage error in radius because area scales with the square of the radius. For a 3% error in radius, the area error is 6%, so the correct option is (A).

    Concept & Intuition

    When a measurement has a small error, the error in a quantity derived from it can be approximated using differentials. For a circle, area $A = \pi r^2$. If the radius has a relative error $\frac{dr}{r}$, then the relative error in area is $\frac{dA}{A} = 2 \frac{dr}{r}$. This is because differentiating $A = \pi r^2$ gives $dA = 2\pi r\, dr$, and dividing by $A = \pi r^2$ yields $\frac{dA}{A} = 2 \frac{dr}{r}$. So the percentage error in area is exactly twice the percentage error in radius — no matter the sign of the error, as long as it’s small.

    Step-by-step solution

    1. Define the given error

       The percentage error in the radius is 3. This means $\frac{\Delta r}{r} \times 100 = 3$, so the relative error in radius is $\frac{\Delta r}{r} = 0.03$.

    2. Relate area to radius

       Area $A = \pi r^2$. For small changes, we use differentials:

    $dA = \frac{dA}{dr} \, dr = 2\pi r \, dr.$

    3. Find the relative error in area Divide $dA$ by $A$:

    $\frac{dA}{A} = \frac{2\pi r \, dr}{\pi r^2} = 2 \frac{dr}{r}.$

    So the relative error in area is twice the relative error in radius. …

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.If the percentage error in the radius of a circle is 3, then the percentage error in its area is (A) $\dfrac{3}{2}$ (B) $2$ (C) $6$ (D) $4$

    ›Reveal solutionSolution

    The percentage error in area is twice the percentage error in the radius because area scales as $r^2$. With a 3% error in radius, the area error is 6%.

    The key idea here is how errors propagate when a quantity depends on a power of another. If you have a measurement with a certain percentage error, any quantity derived from it will have a percentage error that multiplies by the exponent. For a circle, area $A = \pi r^2$, so $A$ is proportional to $r^2$. That means a small relative change in $r$ gets doubled when it becomes a relative change in $A$.

    Let’s work it out step by step.

    1. Write the relation between area and radius.

       The area of a circle is $A = \pi r^2$. Here, $\pi$ is a constant, so any fractional change in $A$ comes entirely from the fractional change in $r$.

    2. Express the percentage error in terms of differentials.

       For small errors, we use the approximation:

    $\frac{\Delta A}{A} \approx \frac{dA}{A} \quad \text{and} \quad \frac{\Delta r}{r} \approx \frac{dr}{r}$

    The percentage error is just $100 \times$ this fractional error.

    3. Differentiate the area formula. Differentiate $A = \pi r^2$ with respect to $r$:

    $\frac{dA}{dr} = 2\pi r$

    So the differential is $dA = 2\pi r \, dr$.

    4. Find the fractional error in area. Divide both sides by $A = \pi r^2$:

    $\frac{dA}{A} = \frac{2\pi r \, dr}{\pi r^2} = 2 \cdot \frac{dr}{r}$

    This is the crucial result: the fractional error in area is exactly twice the fractional error in radius.

    5. Convert to percentage errors. Multiply both sides by 100: $\text{Percentage error in } A = 2 \times (\text{percentage error in } r)$ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The number of common roots among the $12^{\text{th}}$ and $30^{\text{th}}$ roots of unity is (A) $12$ (B) $9$ (C) $8$ (D) $6$

    ›Reveal solutionSolution

    The common roots of the $12^{\text{th}}$ and $30^{\text{th}}$ roots of unity are exactly those roots whose order divides both 12 and 30, i.e., the $\gcd(12,30)=6^{\text{th}}$ roots of unity, so there are 6 common roots.

    The key idea is that the $n^{\text{th}}$ roots of unity are the complex numbers $z$ satisfying $z^n = 1$. A root is common to both the $12^{\text{th}}$ and $30^{\text{th}}$ roots if it satisfies both $z^{12}=1$ and $z^{30}=1$. That means its order must divide both 12 and 30. The set of such roots is exactly the $d^{\text{th}}$ roots of unity where $d = \gcd(12,30)$. Why? Because if a number is a root of both equations, then raising it to any linear combination of 12 and 30 also gives 1; in particular, by Bézout's identity, there exist integers $a,b$ such that $a\cdot 12 + b\cdot 30 = \gcd(12,30)$, so $z^{\gcd}=1$. Conversely, any $d^{\text{th}}$ root of unity automatically satisfies $z^{12}=1$ and $z^{30}=1$ because $d$ divides both exponents. So the common roots are precisely the $d^{\text{th}}$ roots of unity, and there are exactly $d$ of them.

    Let's work through it step by step.

    1. Define the sets.

       The $12^{\text{th}}$ roots of unity are $\{ e^{2\pi i k / 12} \mid k = 0,1,\dots,11 \}$.

       The $30^{\text{th}}$ roots of unity are $\{ e^{2\pi i m / 30} \mid m = 0,1,\dots,29 \}$.

       A common root must be in both sets.

    2. Condition for a common root.

       If $z$ is a common root, then $z^{12}=1$ and $z^{30}=1$. This implies that the order of $z$ (the smallest positive integer $n$ such that $z^n=1$) must divide both 12 and 30. So the order must be a divisor of $\gcd(12,30)$.

    3. Compute the gcd.

       $\gcd(12,30) = 6$. So any common root has an order that divides 6. That means it is a $6^{\text{th}}$ root of unity (or a divisor of 6, but all such are included in the $6^{\text{th}}$ roots).

    4. How many such roots are there? …

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