Q.Find the equation of the circle which touches the both axes in first quadrant and whose radius is $a$.

Showing the 12 most recent of 156 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The centre of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is the point of intersection of the $X$-axis and the line $x + by + g = 0$. If the intercepts made by this circle on $X$ and $Y$ axes are $6$ and $2\sqrt{5}$ respectively, then $g^2 + f^2 + c^2 =$ (A) $41$ (B) $32$ (C) $29$ (D) $25$

   ›Reveal solutionSolution

   Centre-on-$X$-axis fixes $f=0$; the two intercept lengths then pin down $c$ and $g^2$, giving $g^2+f^2+c^2=29$.

   Concept and Intuition

   For a circle $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$. The intercept a circle cuts on the $X$-axis has length $2\sqrt{g^2-c}$, and on the $Y$-axis $2\sqrt{f^2-c}$ — both standard results obtained by setting $y=0$ (or $x=0$) in the circle equation and using the quadratic's root-difference.

   Step-by-Step Solution

   1. Centre is $(-g,-f)$; lying on the $X$-axis ($y=0$) requires $-f=0$, i.e. $f=0$.
   2. The centre must also lie on $x+by+g=0$: substituting $(-g,0)$ gives $-g+0+g=0$, true for any $b$ — so this condition contributes nothing new beyond $f=0$.
   3. $X$-axis intercept length: $2\sqrt{g^2-c}=6\Rightarrow g^2-c=9$.
   4. $Y$-axis intercept length: $2\sqrt{f^2-c}=2\sqrt5\Rightarrow f^2-c=5$. Since $f=0$: $-c=5\Rightarrow c=-5$.
   5. From $g^2-c=9$: $g^2 = 9+c = 9-5=4$.
   6. $g^2+f^2+c^2 = 4+0+25=29$.

   Common Mistakes

   • Trying to solve for $b$ as if it mattered — it drops out entirely once $f=0$ is substituted.
   • Sign slip when solving $f^2-c=5$ with $f=0$ (giving $c=-5$, not $c=5$).

   ✓Final answer

   The correct option is (C) — $29$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The normal drawn at the point $(-1,-2)$ to a circle $S$ is $y + 2 = 0$ and the centre of the circle $S$ does not lie in the $2^{nd}$ and $3^{rd}$ quadrants. If the intercept made by this circle on $X$-axis is $8\sqrt{2}$, then the length of the tangent drawn to the circle from the point $(-1,-1)$ is (A) $1$ (B) $\sqrt{2}$ (C) $3$ (D) $\sqrt{5}$

   ›Reveal solutionSolution

   The horizontal normal fixes the centre's $y$-coordinate; the quadrant restriction and the $X$-intercept pin the centre exactly, giving a tangent length of $1$ from $(-1,-1)$.

   Concept and Intuition

   A normal to a circle at any point always passes through the centre. Here the normal is the horizontal line $y=-2$, so the centre's $y$-coordinate is $-2$. The quadrant restriction then fixes the sign of the centre's $x$-coordinate, and the given point $(-1,-2)$ (through which the normal was drawn) lies on the circle, giving the radius in terms of the unknown $x$-coordinate of the centre.

   Step-by-Step Solution

   1. Normal $y+2=0$ passes through both the point $(-1,-2)$ (given) and the centre, so centre $=(a,-2)$.
   2. Centre not in 2nd quadrant ($x<0,y>0$) is automatic since $y=-2<0$; not in 3rd quadrant ($x<0,y<0$) forces $a\ge0$.
   3. Radius $r =$ distance from centre to the point $(-1,-2)$ on the circle $= |a-(-1)| = a+1$ (positive since $a\ge0$).
   4. The circle's centre is at height $2$ below the $X$-axis, so its $X$-axis intercept length is $2\sqrt{r^2-2^2}=2\sqrt{r^2-4}$. Given this equals $8\sqrt2$: $\sqrt{r^2-4}=4\sqrt2\Rightarrow r^2-4=32\Rightarrow r^2=36\Rightarrow r=6$.
   5. So $a+1=6\Rightarrow a=5$. Centre $=(5,-2)$, radius $=6$.
   6. Tangent length from $(-1,-1)$: $\sqrt{(5-(-1))^2+(-2-(-1))^2-r^2}=\sqrt{36+1-36}=\sqrt1=1$.

   Common Mistakes

   • Forgetting the intercept-length formula needs the perpendicular distance from centre to the axis (here $2$, not $0$).
   • Choosing the wrong sign for $a$ without applying the quadrant restriction.

   ✓Final answer

   The correct option is (A) — $1$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $(h, k)$ is the inverse point of the point $(1, 3)$ with respect to the circle $x^2 + y^2 - 10x + 2y + 1 = 0$, then $h + k =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   The inverse point lies on ray $CP$ scaled so that $CP\cdot CP' = r^2$; direct computation gives $h+k=4$.

   Concept and Intuition

   The inverse of a point $P$ with respect to a circle of centre $C$ and radius $r$ is the point $P'$ on ray $CP$ such that $CP\cdot CP'=r^2$. In vector form, $P' = C + \dfrac{r^2}{|CP|^2}(P-C)$.

   Step-by-Step Solution

   1. Circle $x^2+y^2-10x+2y+1=0$: centre $C=(5,-1)$ (since $2g=-10,2f=2$), radius$^2 = g^2+f^2-c = 25+1-1=25$.
   2. $P=(1,3)$. $P - C = (1-5,\,3-(-1)) = (-4,4)$. $|CP|^2 = 16+16=32$.
   3. $P' = C + \dfrac{25}{32}(-4,4) = (5,-1) + \left(-\dfrac{100}{32},\dfrac{100}{32}\right) = \left(5-\dfrac{25}{8},\,-1+\dfrac{25}{8}\right)$.
   4. $= \left(\dfrac{40-25}{8},\,\dfrac{-8+25}{8}\right) = \left(\dfrac{15}{8},\dfrac{17}{8}\right)$.
   5. $h+k = \dfrac{15}{8}+\dfrac{17}{8} = \dfrac{32}{8}=4$.

   Common Mistakes

   • Confusing the inverse-point formula with the reflection formula (reflection would use $2\times$ projection rather than the $r^2/|CP|^2$ scale factor).
   • Sign errors computing $P-C$.

   ✓Final answer

   The correct option is (D) — $4$.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A circle $x^2 + y^2 - 2x + 2y + 1 = 0$ and a line $x - y - 1 = 0$ intersect at the points A and B. If the equation of the circle having AB as diameter is $x^2 + y^2 + 2gx + 2fy + c = 0$, then $2g + 2f =$ (A) $-2c + 1$ (B) $2c + 2$ (C) $-c + 3$ (D) $c$

   ›Reveal solutionSolution

   The family-of-circles trick plus "centre lies on the diameter-line" pins $\lambda=1$; then $2g+2f=c=0$ identically, matching option (D).

   Concept and Intuition

   Any circle through the two intersection points of a given circle $S=0$ and line $L=0$ can be written as $S+\lambda L=0$. If $AB$ (the common chord) is to be a diameter of this new circle, its centre — being the midpoint of $AB$ — must lie on the line $L$ itself (since $A,B\in L$).

   Step-by-Step Solution

   1. $S: x^2+y^2-2x+2y+1=0$, $L: x-y-1=0$. Family: $x^2+y^2-2x+2y+1+\lambda(x-y-1)=0$, i.e. $x^2+y^2+(\lambda-2)x+(2-\lambda)y+(1-\lambda)=0$.
   2. Centre of this circle: $\left(\dfrac{2-\lambda}{2},\dfrac{\lambda-2}{2}\right)$.
   3. This centre must satisfy $L=0$: $\dfrac{2-\lambda}{2}-\dfrac{\lambda-2}{2}-1=0 \Rightarrow \dfrac{(2-\lambda)-(\lambda-2)}{2}=1\Rightarrow \dfrac{4-2\lambda}{2}=1\Rightarrow 2-\lambda=1\Rightarrow\lambda=1$.
   4. Substituting $\lambda=1$: circle becomes $x^2+y^2-x+y+0=0$. So $2g=-1\Rightarrow g=-\tfrac12$; $2f=1\Rightarrow f=\tfrac12$; $c=0$.
   5. $2g+2f = -1+1=0$, and $c=0$ too — so $2g+2f=c$ holds exactly.

   Common Mistakes

   • Forgetting that "$AB$ is a diameter" translates to a centre-on-line condition, not an equal-radius or other condition.
   • Sign slip in setting up the family of circles equation.

   ✓Final answer

   The correct option is (D) — $c$.

   ANSWER: D

5. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $x - 2y + 3 = 0$ and $x - 5y - 8 = 0$ are the conjugate lines with respect to the circle $x^2 + y^2 - 4x + 6y + c = 0$, then $c =$ (A) $-3$ (B) $-12$ (C) $9$ (D) $4$

   ›Reveal solutionSolution

   Two lines are conjugate w.r.t. a circle when the pole of one lies on the other; solving the pole condition for the given circle and lines gives $c=4$.

   Concept and Intuition

   For a circle $x^2+y^2+2gx+2fy+c=0$, the polar (chord of contact style line) of a point $(x_1,y_1)$ is

   $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0.$

   Two lines $l_1=0$ and $l_2=0$ are called conjugate lines with respect to the circle if the pole of $l_1$ lies on $l_2$ (equivalently the pole of $l_2$ lies on $l_1$ — the relation is symmetric).

   Here the circle is $x^2+y^2-4x+6y+c=0$, so $g=-2,\ f=3$.

   Step-by-Step Solution

   1. Let $(x_1,y_1)$ be the pole of $x-2y+3=0$. The polar of $(x_1,y_1)$ is

   $x(x_1-2)+y(y_1+3)+(-2x_1+3y_1+c)=0.$

   2. Matching this to $x-2y+3=0$ (proportional coefficients), set the common ratio $=k$:

   $x_1-2=k,\quad y_1+3=-2k,\quad -2x_1+3y_1+c=3k.$

   So $x_1=k+2,\ y_1=-2k-3$.

   3. Substitute into the third equation: $-2(k+2)+3(-2k-3)+c=3k \Rightarrow -8k-13+c=3k \Rightarrow c=11k+13$.

   4. The pole $(x_1,y_1)=(k+2,\,-2k-3)$ must lie on the second line $x-5y-8=0$:

   $(k+2)-5(-2k-3)-8=0 \Rightarrow 11k+9=0 \Rightarrow k=-\tfrac{9}{11}.$

   5. Then $c=11\left(-\tfrac{9}{11}\right)+13 = -9+13 = 4$.
   6. (Check by symmetry, computing the pole of the second line and requiring it on the first, also yields $c=4$.)

   Common Mistakes

   • Confusing "conjugate lines" with "conjugate points" (different definitions).
   • Sign errors in writing $g,f$ from $2g=-4,\ 2f=6$.

   ✓Final answer

   The correct option is (D) — $4$.

   ANSWER: D

6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The equation of the circle passing through the points of intersection of the two circles $x^2 + y^2 - 8x - 2y + 16 = 0$, $x^2 + y^2 - 4x - 4y - 1 = 0$ and passing through $(2, -4)$ is (A) $x^2 + y^2 + 104x - 58y - 460 = 0$ (B) $x^2 + y^2 - 104x + 58y - 460 = 0$ (C) $x^2 + y^2 + 52x - 104y + 60 = 0$ (D) $x^2 + y^2 - 104x - 52y - 60 = 0$

   ›Reveal solutionSolution

   Using the family-of-circles-through-intersection trick and plugging in the extra point $(2,-4)$ pins down $\lambda$, giving the circle $x^2+y^2+104x-58y-460=0$.

   Concept and Intuition

   Any circle through the intersection points of $S_1=0$ and $S_2=0$ (both circles with unit $x^2,y^2$ coefficient) can be written as

   $S_1+\lambda(S_2-S_1)=0$

   for some real $\lambda$ — this stays a genuine circle (coefficient of $x^2+y^2$ is still $1$) and automatically passes through both intersection points, since at those points $S_1=S_2=0$.

   Step-by-Step Solution

   1. $S_1: x^2+y^2-8x-2y+16=0$, $\ S_2: x^2+y^2-4x-4y-1=0$.
   2. $S_2-S_1 = (-4x-4y-1)-(-8x-2y+16) = 4x-2y-17$.
   3. Family: $x^2+y^2-8x-2y+16+\lambda(4x-2y-17)=0$.
   4. Substitute $(2,-4)$: $x^2+y^2=4+16=20$, $-8x=-16$, $-2y=8$, constant $16$; sum $=20-16+8+16=28$. Also $4x-2y-17 = 8+8-17=-1$.
   5. So $28-\lambda=0 \Rightarrow \lambda=28$.
   6. Circle: $x^2+y^2-8x-2y+16+28(4x-2y-17)=0 = x^2+y^2+(-8+112)x+(-2-56)y+(16-476)=0$

   $= x^2+y^2+104x-58y-460=0.$

   Common Mistakes

   • Using $S_1+\lambda S_2=0$ (not through-intersection family unless normalized) instead of $S_1+\lambda(S_2-S_1)=0$.
   • Sign slip when distributing $\lambda$ across the linear terms.

   ✓Final answer

   The correct option is (A) — $x^2+y^2+104x-58y-460=0$.

   ANSWER: A

7. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If a circle passing through the points $(1,5)$ and $(4,0)$ makes equal intercepts on coordinate axes and if its centre lies in the first quadrant, then $\sqrt{4g^2 - c^2} =$ (A) $2$ (B) $4$ (C) $3$ (D) $5$

   ›Reveal solutionSolution

   Equal intercepts on the axes force $g^2=f^2$; centre in the first quadrant forces $g=f$ (both negative). Solve using the two given points. Answer: $3$.

   Concept and Intuition

   For circle $x^2+y^2+2gx+2fy+c=0$, the intercept lengths on the axes are $2\sqrt{g^2-c}$ (x-axis) and $2\sqrt{f^2-c}$ (y-axis). Equal intercepts require $g^2=f^2$, i.e. $g=f$ or $g=-f$. The centre is $(-g,-f)$; for this to lie in the first quadrant we need $g<0$ and $f<0$ simultaneously, which is only consistent with $g=f$ (not $g=-f$, which would force opposite signs).

   Step-by-Step Solution

   1. Equal intercepts $\Rightarrow g^2=f^2$; centre $(-g,-f)$ in Q1 $\Rightarrow g<0,\,f<0 \Rightarrow g=f$.
   2. Plug $(1,5)$ into the circle: $1+25+2g(1)+2f(5)+c=0 \Rightarrow 26+2g+10f+c=0$. With $f=g$: $26+12g+c=0 \Rightarrow c=-26-12g$.
   3. Plug $(4,0)$: $16+0+8g+0+c=0 \Rightarrow c=-16-8g$.
   4. Equate: $-26-12g=-16-8g \Rightarrow -10=4g \Rightarrow g=-\dfrac52$.
   5. Then $c=-16-8(-2.5)=-16+20=4$. (Check: centre $(2.5,2.5)$ is in Q1. ✓)
   6. Compute $4g^2-c^2 = 4(6.25)-16 = 25-16=9$, so $\sqrt{4g^2-c^2}=\sqrt9=3$.

   Common Mistakes

   • Taking $g=-f$ instead of $g=f$, which contradicts the first-quadrant centre condition.
   • Sign errors substituting the points into the general circle equation.

   ✓Final answer

   The correct option is (C) — $3$.

   ANSWER: C

8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If a tangent drawn to the circle $x^2 + y^2 - 6x - 8y - 11 = 0$ is perpendicular to the line $3x + 4y + k = 0$ then the distance from the origin to this tangent is (A) $4$ (B) $5$ (C) $6$ (D) $2$

   ›Reveal solutionSolution

   Tangents perpendicular to $3x+4y+k=0$ have the form $4x-3y+c=0$; using the tangency condition on the given circle and then computing distance from the origin gives $6$.

   Concept and Intuition

   Two lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are perpendicular exactly when $a_1a_2+b_1b_2=0$. So any line perpendicular to $3x+4y+k=0$ must have direction coefficients $(4,-3)$ (up to scaling), i.e. it belongs to the family $4x-3y+c=0$. Being tangent to the given circle fixes $c$; then we just measure the origin's distance to that specific line.

   Step-by-Step Solution

   1. The circle $x^2+y^2-6x-8y-11=0$ has centre $C=(3,4)$ and radius $r=\sqrt{3^2+4^2+11}=\sqrt{36}=6$.
   2. Any line perpendicular to $3x+4y+k=0$ is $4x-3y+c=0$ for some constant $c$.
   3. Tangency to the circle means the perpendicular distance from $C$ to this line equals $r$:

   $\frac{|4(3)-3(4)+c|}{\sqrt{4^2+(-3)^2}}=6 \;\Rightarrow\; \frac{|12-12+c|}{5}=6\;\Rightarrow\;|c|=30.$

   4. The distance from the origin $(0,0)$ to $4x-3y+c=0$ is $\dfrac{|c|}{5}=\dfrac{30}{5}=6$.

   Common Mistakes

   • Mixing up the perpendicularity condition with the parallelism condition (parallel lines to $3x+4y+k=0$ would instead be $3x+4y+c=0$).
   • Forgetting that $|c|=30$ gives two possible tangents, but both are equidistant from the origin, so the answer is unambiguous.

   ✓Final answer

   The correct option is (C) — $6$.

   ANSWER: C

9. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $L$ represents a normal drawn at the point $P\left(\dfrac{\pi}{4}\right)$ on the circle $x^2 + y^2 + 6x - 6y - 14 = 0$, then the equation of the diameter of this circle which is perpendicular to $L$ is (A) $x - y + 6 = 0$ (B) $2x + y + 3 = 0$ (C) $3x + 2y + 3 = 0$ (D) $x + y = 0$

   ›Reveal solutionSolution

   The normal at any point of a circle passes through its centre; using the parametric point at $\theta=\pi/4$ gives the normal's slope, and the diameter perpendicular to it (through the centre) works out to $x+y=0$.

   Concept and Intuition

   For a circle, the normal at any boundary point always passes through the centre (radii are normals). So the line $L$ is simply the line joining the centre to $P(\pi/4)$. A "diameter perpendicular to $L$" is just the line through the centre with slope $-1/(\text{slope of }L)$.

   Step-by-Step Solution

   1. Write the circle in standard form: $x^2+6x+y^2-6y=14\Rightarrow (x+3)^2+(y-3)^2=14+9+9=32$. Centre $C=(-3,3)$, radius $r=\sqrt{32}=4\sqrt2$.
   2. Parametrize points on the circle as $(-3+r\cos\theta,\,3+r\sin\theta)$. At $\theta=\pi/4$: $\cos\theta=\sin\theta=\frac{\sqrt2}{2}$, so

   $P=\left(-3+4\sqrt2\cdot\frac{\sqrt2}{2},\;3+4\sqrt2\cdot\frac{\sqrt2}{2}\right)=(-3+4,\,3+4)=(1,7).$

   3. The normal $L$ at $P$ is the line through $C(-3,3)$ and $P(1,7)$: direction $(1-(-3),7-3)=(4,4)$, i.e. slope $1$.
   4. A diameter perpendicular to $L$ passes through the centre with slope $-1$:

   $y-3=-1\cdot(x-(-3))\;\Rightarrow\;y-3=-x-3\;\Rightarrow\;x+y=0.$

   Common Mistakes

   • Forgetting the normal passes through the centre and instead trying to compute a tangent-perpendicular relation.
   • Sign slip when converting $\theta=\pi/4$ to coordinates using the wrong radius.

   ✓Final answer

   The correct option is (D) — $x+y=0$.

   ANSWER: D

10. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If $(h,k)$ is the pole of the line $2x - 3y + 4 = 0$ with respect to the circle $x^2 + y^2 - 4x + 6y - 3 = 0$, then $10h + k =$ (A) $0$ (B) $1$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    Using the polar (chord-of-contact-like) formula $T=0$ for a general circle and matching it proportionally with the given line locates the pole; $10h+k=1$.

    Concept and Intuition

    For a circle $x^2+y^2+2gx+2fy+c=0$, the polar of a point $(x_1,y_1)$ is obtained by the usual $T=0$ substitution: $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. If we're told the polar of an unknown point $(h,k)$ IS a specific line, we just match coefficients (up to a common ratio) between the polar expression and that line.

    Step-by-Step Solution

    1. The circle $x^2+y^2-4x+6y-3=0$ gives $2g=-4\Rightarrow g=-2$, $2f=6\Rightarrow f=3$, $c=-3$.
    2. The polar of $(h,k)$ is $xh+yk+g(x+h)+f(y+k)+c=0$, i.e.

    $x(h+g)+y(k+f)+(gh+fk+c)=0 \;\Rightarrow\; x(h-2)+y(k+3)+(-2h+3k-3)=0.$

    3. This must be the same line as $2x-3y+4=0$, so the coefficients are proportional:

    $\frac{h-2}{2}=\frac{k+3}{-3}=\frac{-2h+3k-3}{4}=t.$

    4. From the first two: $h=2t+2$, $k=-3t-3$. Substitute into the third:

    $-2(2t+2)+3(-3t-3)-3=4t \;\Rightarrow\; -4t-4-9t-9-3=4t \;\Rightarrow\; -13t-16=4t \;\Rightarrow\; t=-\frac{16}{17}.$

    5. So $h=2\left(-\frac{16}{17}\right)+2=\frac{2}{17}$ and $k=-3\left(-\frac{16}{17}\right)-3=-\frac{3}{17}$.
    6. $10h+k=\dfrac{20}{17}-\dfrac{3}{17}=\dfrac{17}{17}=1$.

    Common Mistakes

    • Using the wrong sign convention in the polar formula (mixing up $g,f$ signs from the $2g,2f$ coefficients).
    • Forgetting the proportionality constant $t$ is unknown and must itself be solved for.

    ✓Final answer

    The correct option is (B) — $1$.

    ANSWER: B

11. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If the equation of a circle passing through the point $(2,1)$ and the points of intersection of the circles $x^2 + y^2 + 4x - 6y - 3 = 0$ and $x^2 + y^2 - 2x + 2y - 2 = 0$ is $x^2 + y^2 + 2gx + 2fy + c = 0$, then $2g + f =$ (A) $-2c$ (B) $2c$ (C) $c$ (D) $-c$

    ›Reveal solutionSolution

    Building the family $S_1+\lambda S_2=0$ through the two circles and forcing it through $(2,1)$ pins $\lambda$; comparing coefficients shows $2g+f=c$.

    Concept and Intuition

    Any circle through the intersection points of two circles $S_1=0$ and $S_2=0$ can be written as $S_1+\lambda S_2=0$ for some $\lambda$. Requiring it to also pass through a third given point determines $\lambda$ uniquely (unless that point already lies on $S_2=0$).

    Step-by-Step Solution

    1. $S_1:x^2+y^2+4x-6y-3=0$, $S_2:x^2+y^2-2x+2y-2=0$.
    2. At $(2,1)$: $S_1(2,1)=4+1+8-6-3=4$; $S_2(2,1)=4+1-4+2-2=1$.
    3. Since the required circle passes through $(2,1)$: $S_1+\lambda S_2=0$ there $\Rightarrow 4+\lambda(1)=0\Rightarrow\lambda=-4$.
    4. Substitute into $S_1+\lambda S_2$:

    $(1+\lambda)(x^2+y^2)+(4-2\lambda)x+(-6+2\lambda)y+(-3-2\lambda)=0$

    $\Rightarrow (1-4)(x^2+y^2)+(4+8)x+(-6-8)y+(-3+8)=0 \;\Rightarrow\; -3(x^2+y^2)+12x-14y+5=0.$

    5. Divide by $-3$: $x^2+y^2-4x+\dfrac{14}{3}y-\dfrac53=0$, so $2g=-4\Rightarrow g=-2$, $2f=\dfrac{14}{3}\Rightarrow f=\dfrac73$, $c=-\dfrac53$.
    6. $2g+f=-4+\dfrac73=\dfrac{-12+7}{3}=-\dfrac53=c$.

    Common Mistakes

    • Sign errors when combining $(1+\lambda)$ across all terms — always divide through consistently to bring the $x^2,y^2$ coefficient back to $1$.
    • Mixing up which point's substitution finds $\lambda$ (it must be the extra point, not a point on $S_1$ or $S_2$ alone).

    ✓Final answer

    The correct option is (C) — $c$.

    ANSWER: C

12. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.Equation of the circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length $3a$ is (A) $x^2 + y^2 = 4a^2$ (B) $x^2 + y^2 = 2a^2$ (C) $x^2 + y^2 = 9a^2$ (D) $x^2 + y^2 = 16a^2$

    ›Reveal solutionSolution

    For an equilateral triangle, the circumradius is $\tfrac{2}{3}$ of the median; with median $3a$ this gives radius $2a$, so the circle is $x^2+y^2=4a^2$. Answer: (A).

    Concept and Intuition

    In an equilateral triangle, the median, altitude, angle bisector, and perpendicular bisector from each vertex all coincide, and the centroid, circumcenter, incenter, and orthocentre all coincide at the same point. The centroid always divides a median in the ratio $2:1$ measured from the vertex, so the circumradius (vertex to center) is exactly $\tfrac{2}{3}$ of the median length.

    Step-by-Step Solution

    1. Let the median length be $3a$.
    2. Centroid divides the median $2:1$ from the vertex, so distance from centroid to vertex $=\dfrac{2}{3}\times3a=2a$.
    3. Since the triangle is equilateral, this centroid is also the circumcenter, and $2a$ is the circumradius $R$.
    4. The circle centered at the origin with radius $R=2a$ passing through all three vertices is:

    $x^2+y^2=(2a)^2=4a^2$

    Common Mistakes

    • Confusing the median with the circumradius directly (forgetting the $2:3$ scaling factor).
    • Using the altitude-to-side relation for a general triangle instead of the equilateral-specific centroid ratio.

    ✓Final answer

    The correct option is (A) — $x^2+y^2=4a^2$.

    ANSWER: A