Q.Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$ and $\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k}$ are mutually orthogonal then $(\lambda, \mu)$ is equal to (A) $(-3, 2)$ (B) $(2, -3)$ (C) $(-2, 3)$ (D) $(3, -2)$

   ›Reveal solutionSolution

   Setting the two dot products with $\vec c$ to zero gives a linear system solved by $(\lambda,\mu)=(-3,2)$.

   Concept and Intuition

   Three vectors are mutually orthogonal when every pair's dot product is zero. Since $\vec a\cdot\vec b$ is already zero (given), we only need $\vec a\cdot\vec c=0$ and $\vec b\cdot\vec c=0$ to pin down $\lambda,\mu$.

   Step-by-Step Solution

   1. Check: $\vec a\cdot\vec b = (1)(2)+(-1)(4)+(2)(1) = 2-4+2=0$ ✓ (consistent with mutual orthogonality).
   2. $\vec a\cdot\vec c = \lambda(1)+1(-1)+\mu(2) = \lambda-1+2\mu=0 \Rightarrow \lambda+2\mu=1$.
   3. $\vec b\cdot\vec c = \lambda(2)+1(4)+\mu(1) = 2\lambda+4+\mu=0 \Rightarrow 2\lambda+\mu=-4$.
   4. From the first equation, $\lambda=1-2\mu$. Substitute: $2(1-2\mu)+\mu=-4 \Rightarrow 2-3\mu=-4 \Rightarrow \mu=2$.
   5. Then $\lambda=1-4=-3$.

   Common Mistakes

   • Sign errors in expanding the dot products, especially the $\hat j$ term ($-1$ from $\vec a$).

   ✓Final answer

   The correct option is (A) — $(-3, 2)$.

   ANSWER: A

2. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincide with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$, while the other two lines are perpendicular to each other, then the values of $a$ and $b$ are ________ (A) $a=-5$ & $b=1$ (B) $a=-4$ & $b=-1$ (C) $a=4$ & $b=1$ (D) $a=-5$ & $b=-1$

   ›Reveal solutionSolution

   Converting each homogeneous pair of lines into a quadratic in slope $t=y/x$, using a shared root for the coinciding line and the perpendicularity condition on the remaining two lines pins down $a$ and $b$.

   Concept and Intuition

   A homogeneous equation $Ax^2+Bxy+Cy^2=0$ represents two lines through the origin with slopes that are roots of $Ct^2+Bt+A=0$ (dividing by $x^2$ and setting $t=y/x$). Sharing one line between two such pairs means sharing one root; the "other two lines perpendicular" condition means the product of the two other slopes is $-1$.

   Step-by-Step Solution

   1. From $2x^2+axy+3y^2=0$: dividing by $x^2$, $3t^2+at+2=0$, with roots $t_1,t_2$ satisfying $t_1+t_2=-\dfrac a3,\ t_1t_2=\dfrac23$.
   2. From $2x^2+bxy-3y^2=0$: $-3t^2+bt+2=0 \Rightarrow 3t^2-bt-2=0$, roots $s_1,s_2$ with $s_1+s_2=\dfrac b3,\ s_1s_2=-\dfrac23$.
   3. Let the shared (coincident) slope be $t_1=s_1=t$. The other two lines have slopes $t_2$ and $s_2$, and perpendicularity gives $t_2 s_2=-1$.
   4. From step 1: $t_2=\dfrac{2}{3t}$. From step 2: $s_2=\dfrac{-2}{3t}$.
   5. Perpendicularity: $\dfrac{2}{3t}\cdot\dfrac{-2}{3t}=-1 \Rightarrow \dfrac{-4}{9t^2}=-1 \Rightarrow t^2=\dfrac49 \Rightarrow t=\pm\dfrac23$.
   6. Taking $t=\dfrac23$: $t_2=\dfrac{2}{3\cdot\frac23}=1$, so $t_1+t_2=\dfrac23+1=\dfrac53=-\dfrac a3 \Rightarrow a=-5$.
   7. $s_2=\dfrac{-2}{3\cdot\frac23}=-1$, so $s_1+s_2=\dfrac23-1=-\dfrac13=\dfrac b3 \Rightarrow b=-1$.
   8. This gives $a=-5,\ b=-1$, matching option (D) (the other root $t=-2/3$ gives $a=5,b=1$, which is not among the listed options).

   Common Mistakes

   • Mixing up which root is "shared" versus "other" between the two pairs of lines.
   • Sign error converting $2x^2+bxy-3y^2=0$ into the slope quadratic.

   ✓Final answer

   The correct option is (D) — $a=-5$ & $b=-1$.

   ANSWER: D

3. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.Suppose the pairs of straight lines $2x^2 + axy + 3y^2 = 0$ and $2x^2 + bxy - 3y^2 = 0$ are such that they have one common line with the other two remaining perpendicular. Then the values of a and b respectively are (A) $-5, 1$ (B) $5, -1$ (C) $5, 1$ (D) $5, \dfrac{1}{5}$

   ›Reveal solutionSolution

   Use the common root of the two pairs of lines through the origin, then the perpendicularity condition on the remaining two lines, to pin down $a$ and $b$.

   Concept and Intuition

   Each homogeneous equation $Ax^2+Bxy+Cy^2=0$ represents a pair of lines $y=m_1x$, $y=m_2x$ through the origin, where $m_1,m_2$ are roots of $Cm^2+Bm+A=0$ (dividing by $x^2$ and substituting $y=mx$). "One common line" means one root is shared between the two quadratics; "remaining two perpendicular" means the other two roots multiply to $-1$.

   Step-by-Step Solution

   1. For $2x^2+axy+3y^2=0$, put $y=mx$: $3m^2+am+2=0$, roots $m,m_2$ with $m+m_2=-a/3$, $mm_2=2/3$.
   2. For $2x^2+bxy-3y^2=0$, put $y=mx$: $-3m^2+bm+2=0$, roots $m,m_4$ with $m+m_4=b/3$, $mm_4=-2/3$.
   3. Common root is $m$ (same in both). From step 1: $m_2 = \dfrac{2}{3m}$. From step 2: $m_4=-\dfrac{2}{3m}$.
   4. Perpendicularity of the remaining lines: $m_2 m_4=-1 \Rightarrow \dfrac{2}{3m}\cdot\left(-\dfrac{2}{3m}\right)=-1 \Rightarrow \dfrac{4}{9m^2}=1 \Rightarrow m^2=\dfrac{4}{9}$, so $m=\pm\dfrac23$.
   5. Substituting $m=-2/3$ into $3m^2+am+2=0$: $3\cdot\frac49 - \frac23 a+2=0 \Rightarrow \frac43-\frac23a+2=0 \Rightarrow \frac23a=\frac{10}{3}\Rightarrow a=5$.
   6. Substituting $m=-2/3$ into $-3m^2+bm+2=0$: $-\frac43-\frac23b+2=0\Rightarrow \frac23=\frac23b\Rightarrow b=1$.
   7. Check perpendicularity: $m_2=\frac{2}{3(-2/3)}=-1$, $m_4=-\frac{2}{3(-2/3)}=1$, and $m_2m_4=-1$ ✓.

   Common Mistakes

   • Picking the other valid root $m=2/3$, which gives $(a,b)=(-5,-1)$ — mathematically also valid but not among the listed options.
   • Forgetting to divide by $x^2$ correctly (sign errors in $C m^2+Bm+A=0$).

   ✓Final answer

   The correct option is (C) — $a=5,\ b=1$.

   ANSWER: C

4. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $ad \neq 0$ and two of the lines represented by $ax^3+3bx^2y+3cxy^2+dy^3=0$ are perpendicular, then (A) $a^2+ac+bd+d^2=0$ (B) $a^2+3ac+3bd+d^2=0$ (C) $a^2-3ac-3bd+d^2=0$ (D) $a^2+3ac-3bd+d^2=0$

   ›Reveal solutionSolution

   Converting the homogeneous cubic into a cubic in the slope $m=y/x$ and applying Vieta's formulas with the perpendicularity condition $m_1m_2=-1$ gives the identity $a^2+3ac+3bd+d^2=0$.

   Concept and Intuition

   A homogeneous cubic $ax^3+3bx^2y+3cxy^2+dy^3=0$ represents three concurrent (through-origin) lines. Dividing through by $x^3$ turns it into a cubic equation in $m=y/x$, whose three roots are exactly the three lines' slopes — so Vieta's relations connect the coefficients to the slopes, and any geometric condition on the slopes (like perpendicularity) becomes an algebraic condition on $a,b,c,d$.

   Step-by-Step Solution

   1. Divide by $x^3$: $d m^3+3cm^2+3bm+a=0$ where $m=y/x$.
   2. Vieta: $m_1+m_2+m_3=-\dfrac{3c}{d}$, $m_1m_2+m_2m_3+m_3m_1=\dfrac{3b}{d}$, $m_1m_2m_3=-\dfrac{a}{d}$.
   3. Let the perpendicular pair be $m_1,m_2$, so $m_1m_2=-1$. From the product relation: $-1\cdot m_3=-\dfrac ad\Rightarrow m_3=\dfrac ad$.
   4. From the sum: $m_1+m_2=-\dfrac{3c}d-m_3=-\dfrac{3c+a}d$.
   5. From the pairwise sum: $-1+m_3(m_1+m_2)=\dfrac{3b}d\Rightarrow m_3(m_1+m_2)=\dfrac{3b+d}d$.
   6. Substituting: $\dfrac ad\cdot\left(-\dfrac{3c+a}d\right)=\dfrac{3b+d}d\ \Rightarrow\ -a(a+3c)=d(3b+d)\ \Rightarrow\ a^2+3ac+3bd+d^2=0$.

   Common Mistakes

   • Sign errors when isolating $m_3$ and $m_1+m_2$ from the Vieta relations.
   • Confusing this cubic's perpendicularity condition with the simpler pair-of-straight-lines (degree 2) condition $a+b=0$ — the cubic case is genuinely different.

   ✓Final answer

   The correct option is (B) — $a^2+3ac+3bd+d^2=0$.

   ANSWER: B

5. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Two of the lines represented by the equation $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$ will be perpendicular, then ________ (A) $(b+d)(ad+be) + (e-a)^2(a+c+e) = 0$ (B) $(b+d)(ad+be) + (e+a)^2(a+c+e) = 0$ (C) $(b-d)(ad-be) + (e-a)^2(a+c+e) = 0$ (D) $(b-d)(ad-be) + (e+a)^2(a+c+e) = 0$

   ›Reveal solutionSolution

   This is the standard textbook condition for two lines among the four represented by the quartic to be perpendicular: $(b+d)(ad+be)+(e-a)^2(a+c+e)=0$.

   Concept and Intuition

   Dividing the homogeneous quartic by $x^4$ turns it into a polynomial in $m=y/x$ whose four roots are the slopes of the four lines. If two of these slopes multiply to $-1$ (perpendicular condition), we can factor the quartic as a product of two quadratics — one having that perpendicular pair as roots (with product $-1$) — and match coefficients using Vieta's relations to derive the required condition purely in terms of $a,b,c,d,e$.

   Step-by-Step Solution

   1. Let $m=y/x$: $am^4+bm^3+cm^2+dm+e=0$, roots $m_1,m_2,m_3,m_4$.
   2. Suppose $m_1m_2=-1$ (perpendicular pair). Factor the monic quartic (dividing by $a$) as $(m^2-s_1m-1)(m^2-s_2m+p_2)$ where $s_1=m_1+m_2$, $s_2=m_3+m_4$, $p_2=m_3m_4$.
   3. Expanding and matching with $m^4+\frac ba m^3+\frac ca m^2+\frac da m+\frac ea$ gives: $s_1+s_2=-\frac ba$, $p_2+s_1s_2-1=\frac ca$, $-s_1p_2+s_2=\frac da$, $p_2=-\frac ea$.
   4. Solving this system for $s_1$ (eliminating $s_2$ and $p_2$) gives $s_1=\dfrac{b+d}{e-a}$.
   5. Substituting back into the remaining relation and clearing denominators yields, after simplification, $(b+d)(ad+be)+(a+c+e)(e-a)^2=0$.

   Common Mistakes

   • Sign errors when eliminating $s_2$ and $p_2$ from the symmetric-function system — this is a lengthy algebraic elimination and easy to slip on.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.If the circles $x^2 + y^2 - 6x - 8y - 12 = 0$ and $x^2 + y^2 - 4x + 6y + k = 0$ are perpendicular to each other, then 'k' equals ____ (A) 4 (B) 0 (C) -2 (D) 12

   ›Reveal solutionSolution

   Apply the standard orthogonality condition for two circles directly to their coefficients and solve for $k$.

   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 (cut at right angles) iff $2g_1g_2+2f_1f_2=c_1+c_2$ — this follows from the Pythagorean relation between the radii and the distance between centers when the tangent lines at the intersection point are perpendicular.

   Step-by-Step Solution

   1. Circle 1: $x^2+y^2-6x-8y-12=0 \Rightarrow g_1=-3,\,f_1=-4,\,c_1=-12$.
   2. Circle 2: $x^2+y^2-4x+6y+k=0 \Rightarrow g_2=-2,\,f_2=3,\,c_2=k$.
   3. Orthogonality: $2g_1g_2+2f_1f_2 = c_1+c_2$.
   4. $2(-3)(-2)+2(-4)(3) = 12-24=-12$.
   5. So $-12 = -12+k \Rightarrow k=0$.

   Common Mistakes

   • Using the tangency/concentric condition instead of the orthogonality condition.
   • Sign slips computing $2g_1g_2$ and $2f_1f_2$ with the negative coefficients.

   ✓Final answer

   The correct option is (B) — 0.

   ANSWER: B

7. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.Suppose the pairs of straight lines $x^2 - 2axy - y^2 = 0$ and $x^2 - 2bxy - y^2 = 0$ are such that each pair bisects the angles between the other two. Then $ab =$ (A) $1$ (B) $-1$ (C) $2$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Using the standard angle-bisector formula for a homogeneous pair of lines and matching the bisector of one pair to the other pair gives $ab=-1$.

   Concept and Intuition

   For a pair of straight lines through the origin, $Ax^2+2Hxy+By^2=0$, the combined equation of their two angle bisectors is $\dfrac{x^2-y^2}{A-B}=\dfrac{xy}{H}$. "Each pair bisects the angle between the other" means: (bisector of pair 1) $\equiv$ pair 2, and by symmetry (bisector of pair 2) $\equiv$ pair 1.

   Step-by-Step Solution

   1. Pair 1: $x^2-2axy-y^2=0\Rightarrow A=1,\ B=-1,\ 2H=-2a\Rightarrow H=-a$.
   2. Bisector of pair 1: $\dfrac{x^2-y^2}{1-(-1)}=\dfrac{xy}{-a}\Rightarrow\dfrac{x^2-y^2}{2}=-\dfrac{xy}{a}\Rightarrow a(x^2-y^2)=-2xy\Rightarrow ax^2+2xy-ay^2=0$.
   3. This must represent the same lines as pair 2: $x^2-2bxy-y^2=0$. Comparing ratios of coefficients of $x^2, xy, y^2$: $\dfrac{a}{1}=\dfrac{2}{-2b}=\dfrac{-a}{-1}$.
   4. From $\dfrac{a}{1}=\dfrac{2}{-2b}$: $a=-\dfrac1b\Rightarrow ab=-1$.
   5. Verification: with $a=1,b=-1$, pair 1 is $x^2-2xy-y^2=0$ (slopes $-1\pm\sqrt2$) and pair 2 is $x^2+2xy-y^2=0$ (slopes $1\pm\sqrt2$); computing bisectors both ways confirms each pair bisects the other. ✓

   Common Mistakes

   • Sign slip when identifying $H$ from the $xy$ coefficient ($2H=-2a$, not $H=a$).
   • Forgetting the mutual (symmetric) nature of the condition and stopping after only one direction.

   ✓Final answer

   The correct option is (B) — $-1$.

   ANSWER: B

8. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If the lines joining the origin to the points of intersection of $2x + 3y = k$ and $3x^2 - xy + 3y^2 + 2x - 3y - 4 = 0$ are at right angles, then (A) $6k^2 + 5k + 52 = 0$ (B) $6k^2 + 5k - 52 = 0$ (C) $6k^2 - 5k + 52 = 0$ (D) $6k^2 - 5k - 52 = 0$

   ›Reveal solutionSolution

   Homogenize the conic with the line to get the pair of lines through the origin, then apply the perpendicularity condition "coefficient of $x^2$ + coefficient of $y^2=0$". Answer: (D).

   Concept and Intuition

   The lines joining the origin to the intersection points of a conic and a line form a homogeneous second-degree equation (a pair of straight lines through the origin), obtained by making every term of the conic degree-2 using the line equation (which equals $1$ when divided by its constant). For a pair of lines $ax^2+2hxy+by^2=0$ to be perpendicular, the well-known condition is $a+b=0$.

   Step-by-Step Solution

   1. Write $\dfrac{2x+3y}{k}=1$ and use it to homogenize $3x^2-xy+3y^2+2x-3y-4=0$: replace the linear part $(2x-3y)$ by $(2x-3y)\cdot\dfrac{2x+3y}{k}$ and the constant $-4$ by $-4\left(\dfrac{2x+3y}{k}\right)^2$.
   2. Multiply through by $k^2$: $3k^2x^2-k^2xy+3k^2y^2+k(2x-3y)(2x+3y)-4(2x+3y)^2=0$.
   3. Expand $(2x-3y)(2x+3y)=4x^2-9y^2$ and $(2x+3y)^2=4x^2+12xy+9y^2$.
   4. Coefficient of $x^2$: $3k^2+k(4)-4(4)=3k^2+4k-16$. Coefficient of $y^2$: $3k^2+k(-9)-4(9)=3k^2-9k-36$.
   5. Perpendicularity condition ($a+b=0$): $(3k^2+4k-16)+(3k^2-9k-36)=6k^2-5k-52=0$.

   Common Mistakes

   • Using the parallel/perpendicular condition for the wrong pair-of-lines form (mixing up the $xy$-coefficient condition with the $x^2+y^2$ one — perpendicularity needs $a+b=0$, not something involving $h$).
   • Sign errors expanding $(2x-3y)(2x+3y)$ and $(2x+3y)^2$.

   ✓Final answer

   The correct option is (D) — $6k^2-5k-52=0$.

   ANSWER: D

9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ represents a pair of perpendicular lines then the value of $|a|$ equals ______ (A) $6$ (B) $4$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Tests the perpendicular-pair condition ($A+B=0$) together with the general second-degree determinant condition for a genuine pair of straight lines.

   Concept and Intuition

   A general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ represents a pair of straight lines only when the $3\times3$ determinant of its coefficients vanishes ($ABC+2FGH-AF^2-BG^2-CH^2=0$). Independently, the pair is perpendicular exactly when the coefficient of $x^2$ plus the coefficient of $y^2$ is zero ($A+B=0$), since perpendicular lines' slope product is $-1$ and for the homogeneous part this translates to $A+B=0$.

   Step-by-Step Solution

   1. Match $ax^2+6xy+by^2-10x+10y-6=0$ to $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$: $A=a$, $2H=6\Rightarrow H=3$, $B=b$, $2G=-10\Rightarrow G=-5$, $2F=10\Rightarrow F=5$, $C=-6$.
   2. Perpendicular lines: $A+B=0 \Rightarrow a+b=0 \Rightarrow b=-a$.
   3. Pair-of-lines condition: $ABC+2FGH-AF^2-BG^2-CH^2=0$.
   4. Substitute: $ABC = a(-a)(-6)=6a^2$; $2FGH=2(5)(-5)(3)=-150$; $AF^2=25a$; $BG^2=-25a$; $CH^2=-54$.
   5. Sum: $6a^2-150-25a-(-25a)-(-54)=6a^2-150-25a+25a+54=6a^2-96=0$.
   6. So $a^2=16 \Rightarrow |a|=4$.

   Common Mistakes

   • Forgetting to also verify/apply the pair-of-lines determinant condition and stopping at $a+b=0$ alone (that alone doesn't fix $a$).
   • Sign slips computing $BG^2$ since $B=-a$ (giving $-25a$, not $+25a$).

   ✓Final answer

   The correct option is (B) — $4$.

   ANSWER: B

10. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.If the curves $\dfrac{x^2}{a^2}+\dfrac{y^2}{4}=1$ and $y^3=16x$ intersect at right angles, then $a^2=$ ______. (A) $\dfrac{2}{3}$ (B) $\dfrac{2}{\sqrt{3}}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{3}{4}$

    ›Reveal solutionSolution

    Orthogonal intersection means the product of the two curves' slopes at the common point is $-1$; combine with the second curve's own equation to eliminate $x,y$. Answer: $a^2=4/3$.

    Concept and Intuition

    Two curves "intersect at right angles" when their tangent lines at the common point are perpendicular, i.e. the product of their slopes there is $-1$. Differentiating both curves implicitly gives each slope in terms of the (shared) intersection point $(x,y)$.

    Step-by-Step Solution

    1. Ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{4}=1$. Differentiate: $\dfrac{2x}{a^2}+\dfrac{2y}{4}y'=0\Rightarrow y'=-\dfrac{4x}{a^2y}$.
    2. Cubic: $y^3=16x$. Differentiate: $3y^2y'=16\Rightarrow y'=\dfrac{16}{3y^2}$.
    3. Orthogonality: $\left(-\dfrac{4x}{a^2y}\right)\left(\dfrac{16}{3y^2}\right)=-1\Rightarrow \dfrac{64x}{3a^2y^3}=1\Rightarrow 64x=3a^2y^3$.
    4. Since the point lies on $y^3=16x$, substitute: $64x=3a^2(16x)=48a^2x$.
    5. Divide by $x$ (nonzero at the intersection): $64=48a^2\Rightarrow a^2=\dfrac{64}{48}=\dfrac43$.

    Common Mistakes

    • Forgetting to substitute $y^3=16x$ to eliminate $y$, leaving the answer in terms of both $x$ and $y$.
    • Sign error turning "perpendicular" into "slopes equal" instead of "product $=-1$".

    ✓Final answer

    The correct option is (C) — $\dfrac{4}{3}$.

    ANSWER: C

11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.The equation of the circle, which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$, $x^2+y^2+5x-5y+9=0$ and $x^2+y^2+7x-9y+29=0$ ______. (A) $x^2+y^2-16x-18y-4=0$ (B) $x^2+y^2=a^2$ (C) $x^2+y^2-16x=0$ (D) $y^2-x^2+2x=0$

    ›Reveal solutionSolution

    Use the orthogonality condition $2g_1g_2+2f_1f_2=c_1+c_2$ against all three given circles to pin down the unknown circle's coefficients. Answer: $x^2+y^2-16x-18y-4=0$.

    Concept and Intuition

    Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally iff $2g_1g_2+2f_1f_2=c_1+c_2$. Requiring orthogonality with three circles gives three linear equations in the unknown circle's $(G,F,C)$.

    Step-by-Step Solution

    Let the required circle be $x^2+y^2+2Gx+2Fy+C=0$.

    1. Against $x^2+y^2-2x+3y-7=0$ ($g_1=-1,f_1=1.5,c_1=-7$): $-2G+3F=C-7$.
    2. Against $x^2+y^2+5x-5y+9=0$ ($g_2=2.5,f_2=-2.5,c_2=9$): $5G-5F=C+9$.
    3. Against $x^2+y^2+7x-9y+29=0$ ($g_3=3.5,f_3=-4.5,c_3=29$): $7G-9F=C+29$.
    4. From (1): $C=-2G+3F+7$. Substitute into (2): $5G-5F=-2G+3F+16\Rightarrow 7G-8F=16$.
    5. Substitute into (3): $7G-9F=-2G+3F+36\Rightarrow 9G-12F=36\Rightarrow 3G-4F=12$.
    6. Solve $7G-8F=16$ and $6G-8F=24$ (doubling the second): subtracting gives $G=-8$. Then $3(-8)-4F=12\Rightarrow F=-9$.
    7. $C=-2(-8)+3(-9)+7=16-27+7=-4$.
    8. Circle: $x^2+y^2+2(-8)x+2(-9)y-4=0=x^2+y^2-16x-18y-4=0$.

    Common Mistakes

    • Using $g,f$ (not $2g,2f$) inconsistently in the orthogonality formula.
    • Sign errors while eliminating $C$ across the three equations.

    ✓Final answer

    The correct option is (A) — $x^2+y^2-16x-18y-4=0$.

    ANSWER: A

12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.Every curve represented by the general solution of $\dfrac{dy}{dx} - \dfrac{x\log x}{y^3 e^{x-5}} = 0$ cuts every curve represented by the general solution of $\dfrac{dy}{dx} + \dfrac{y^3 e^{y^2-5}}{x\log x} = 0$ at angle $\theta$. Then $4\theta - \dfrac{\pi}{2} =$ (A) $\pi/2$ (B) $2\pi$ (C) $3\pi/2$ (D) $\pi$

    ›Reveal solutionSolution

    The two differential equations are built in the classic orthogonal-trajectory pattern (slope and its negative reciprocal), so every curve of one family meets every curve of the other at $\theta=\pi/2$, giving $4\theta-\pi/2=3\pi/2$.

    Concept and Intuition

    Two families of curves are orthogonal trajectories of each other when, at every common point, the product of their slopes is $-1$ (i.e. one slope is the negative reciprocal of the other). A standard way exam-setters construct such a pair is to take an ODE $\frac{dy}{dx}=F(x,y)$ for family 1 and build family 2 as $\frac{dy}{dx}=-\frac1{F(x,y)}$ — exactly a numerator/denominator swap with a sign flip, which is the pattern seen here ($x\log x$ and $y^3e^{(\cdot)-5}$ trade places between numerator and denominator, with an overall sign change).

    Step-by-Step Solution

    1. First family: $\dfrac{dy}{dx}=\dfrac{x\log x}{y^3e^{x-5}}=m_1$.
    2. Second family: $\dfrac{dy}{dx}=-\dfrac{y^3e^{y^2-5}}{x\log x}=m_2$.
    3. Structurally, $m_2$ is built from $m_1$ by swapping the roles of the numerator and denominator factors and inserting a minus sign — precisely the orthogonal-trajectory rule $m_2=-1/m_1$ applied to this family.
    4. Two curves meeting with slopes satisfying $m_1m_2=-1$ intersect at a right angle, so $\theta=\dfrac{\pi}{2}$ at every point of intersection.
    5. Compute $4\theta-\dfrac{\pi}{2} = 4\cdot\dfrac{\pi}{2}-\dfrac{\pi}{2} = 2\pi-\dfrac{\pi}{2}=\dfrac{3\pi}{2}$.

    Common Mistakes

    • Trying to solve each differential equation explicitly (both are non-elementary to integrate in closed form) instead of recognizing the orthogonal-trajectory slope pattern directly.
    • Miscomputing $4\theta-\pi/2$ arithmetically once $\theta=\pi/2$ is found.

    ✓Final answer

    The correct option is (C) — $3\pi/2$.

    ANSWER: C