Q.A line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$. Obtain its equation.

Showing the 12 most recent of 59 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the line $4x-3y+c=0$ $(c<-10)$ makes an intercept of length $10$ units on the circle $x^2+y^2-2x+4y-23=0$, then $c=$ (A) $-10-\sqrt{175}$ (B) $-10-\sqrt{75}$ (C) $-10-\sqrt{125}$ (D) $-10-\sqrt{150}$

   ›Reveal solutionSolution

   The key idea is to use the chord-length formula for a circle: the length of the intercept is $2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from the center to the line. Solving for $c$ with the given condition $c < -10$ yields $c = -10 - \sqrt{125}$, which corresponds to option (C).

   ---

   We start with the circle equation:

   $x^2 + y^2 - 2x + 4y - 23 = 0$

   Complete the square for $x$ and $y$:

   • For $x$: $x^2 - 2x = (x-1)^2 - 1$
   • For $y$: $y^2 + 4y = (y+2)^2 - 4$

   So the circle becomes:

   $(x-1)^2 + (y+2)^2 - 1 - 4 - 23 = 0 \quad\Rightarrow\quad (x-1)^2 + (y+2)^2 = 28$

   Thus the center is $C(1, -2)$ and radius $r = \sqrt{28} = 2\sqrt{7}$.

   ---

   The line is:

   $4x - 3y + c = 0$

   The perpendicular distance from the center $(1, -2)$ to this line is:

   $d = \frac{|4(1) - 3(-2) + c|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 + 6 + c|}{5} = \frac{|c + 10|}{5}$

   ---

   The chord (intercept) length on the circle is given by:

   $\text{Length} = 2\sqrt{r^2 - d^2}$

   We are told this length is $10$ units, so:

   $2\sqrt{r^2 - d^2} = 10 \quad\Rightarrow\quad \sqrt{r^2 - d^2} = 5 \quad\Rightarrow\quad r^2 - d^2 = 25$

   Substitute $r^2 = 28$:

   $28 - d^2 = 25 \quad\Rightarrow\quad d^2 = 3 \quad\Rightarrow\quad d = \sqrt{3}$

   (We take positive distance.)

   ---

   Now $d = \frac{|c+10|}{5} = \sqrt{3}$, so:

   $|c+10| = 5\sqrt{3}$

   Thus:

   $c + 10 = \pm 5\sqrt{3} \quad\Rightarrow\quad c = -10 \pm 5\sqrt{3}$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.$y = 4$ is the directrix of the parabola $x^2 + 8x + 12y + k = 0$. If $l$ is the length of its latus rectum, then $l - k =$ (A) $4$ (B) $8$ (C) $12$ (D) $6$

   ›Reveal solutionSolution

   The key idea is to rewrite the given equation in standard parabola form, extract the focal length from the directrix condition, compute the latus rectum length $l$, and then find $l - k$. The final result is $l - k = 8$.

   We start with the equation of the parabola:

   $x^2 + 8x + 12y + k = 0$

   Our goal is to find $l$ (length of latus rectum) and $k$ (the constant term), then compute $l - k$.

   ---

   Concept and Intuition

   A parabola with a vertical axis has the standard form $(x - h)^2 = 4p(y - k)$, where $p$ is the focal length (distance from vertex to focus, and also from vertex to directrix). The directrix is a horizontal line $y = k - p$. The latus rectum length is $|4p|$. Here we are told the directrix is $y = 4$, so we can match forms to find $p$ and the vertex, then recover $k$.

   ---

   Step-by-step solution

   1. Complete the square in $x$ The given equation is:

   $x^2 + 8x + 12y + k = 0$

   Group the $x$-terms:

   $(x^2 + 8x) + 12y + k = 0$

   Complete the square: $x^2 + 8x = (x+4)^2 - 16$.

   Substitute:

   $(x+4)^2 - 16 + 12y + k = 0$

   $(x+4)^2 + 12y + (k - 16) = 0$

   2. Isolate the squared term Move the $y$-term and constant to the right:

   $(x+4)^2 = -12y - (k - 16)$

   $(x+4)^2 = -12y - k + 16$

   Factor $-12$ from the right-hand side to match the standard form $(x-h)^2 = 4p(y - \text{vertex}_y)$:

   $(x+4)^2 = -12\left(y + \frac{k - 16}{12}\right)$

   3. Identify the vertex and focal length Compare with $(x - h)^2 = 4p(y - v)$, where $(h, v)$ is the vertex. Here $h = -4$, and $4p = -12$, so:

   $p = -3$

   The vertex’s $y$-coordinate is:

   $v = -\frac{k - 16}{12}$

   4. Use the directrix condition For a vertical parabola $(x-h)^2 = 4p(y - v)$, the directrix is:

   $y = v - p$

   We are told the directrix is $y = 4$. So:

   $v - p = 4$

   Substitute $p = -3$ and $v = -\frac{k - 16}{12}$:

   $$$$

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The area of the rectangle formed by the tangents drawn at the ends of both major and minor axes of an ellipse is 24. If the eccentricity of the ellipse is $\dfrac{1}{4}$, then the equation of the ellipse is (A) $\dfrac{x^2}{48} + \dfrac{y^2}{45} = 1$ (B) $\dfrac{x^2}{16} + \dfrac{y^2}{15} = 1$ (C) $\dfrac{x^2}{24} + \dfrac{y^2}{1/\sqrt{5}} = 1$ (D) $\dfrac{x^2}{8\sqrt{3}} + \dfrac{2y^2}{15\sqrt{3}} = \dfrac{1}{\sqrt{5}}$

   ›Reveal solutionSolution

   The tangent rectangle has sides $2a,2b$, so $4ab=24\Rightarrow ab=6$; with $e=\tfrac14$, $b^2=\tfrac{15}{16}a^2$, giving $a^2=\tfrac{8\sqrt{15}}{5}$, $b^2=\tfrac{3\sqrt{15}}{2}$ — option (D).

   Tangent rectangle. Tangents at the axis-ends are $x=\pm a$ and $y=\pm b$, forming a rectangle of sides $2a$ and $2b$. Area $=4ab=24\Rightarrow ab=6\Rightarrow a^2b^2=36.$

   Eccentricity. $e^2=1-\dfrac{b^2}{a^2}=\dfrac1{16}\Rightarrow \dfrac{b^2}{a^2}=\dfrac{15}{16}\Rightarrow b^2=\dfrac{15}{16}a^2.$

   Solve. $a^2b^2=a^2\cdot\dfrac{15}{16}a^2=\dfrac{15}{16}a^4=36\Rightarrow a^4=\dfrac{576}{15}\Rightarrow a^2=\dfrac{24}{\sqrt{15}}=\dfrac{8\sqrt{15}}{5}.$ Then $b^2=\dfrac{15}{16}\cdot\dfrac{8\sqrt{15}}{5}=\dfrac{3\sqrt{15}}{2}.$ …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the distance of a variable point P from a fixed point $(3, -4)$ is $\dfrac{2}{3}$ times the distance of P from a fixed line $x-y+2=0$ and the locus of that point P is $ax^2+4xy+by^2-62x+80y+c=0$ then $2c=$ (A) $31ab$ (B) $31(a+b)$ (C) $7ab$ (D) $7(a+b)$

   ›Reveal solutionSolution

   The locus reduces to $7x^2+4xy+7y^2-62x+80y+217=0$, so $a=b=7,\ c=217$ and $2c=434=31(a+b)$.

   Let $P=(x,y)$. The condition is

   $\sqrt{(x-3)^2+(y+4)^2}=\frac23\cdot\frac{|x-y+2|}{\sqrt2}.$

   Squaring both sides:

   $(x-3)^2+(y+4)^2=\frac{4}{9}\cdot\frac{(x-y+2)^2}{2}=\frac{2}{9}(x-y+2)^2.$

   Multiply through by $9$:

   $9\left[(x-3)^2+(y+4)^2\right]=2(x-y+2)^2.$

   Expanding each side,

   $9x^2+9y^2-54x+72y+225=2x^2+2y^2-4xy+8x-8y+8.$ …

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the equation of a transverse common tangent drawn to the circles $x^2 + y^2 - 2x - 10y + 1 = 0$ and $x^2 + y^2 + 8x + 14y + 1 = 0$ is $5x + by + c = 0$ then $b + c =$ (A) $17$ (B) $7$ (C) $12$ (D) $24$

   ›Reveal solutionSolution

   The problem asks for the sum $b+c$ in the equation of a transverse common tangent to two circles. The key is to find the tangent that separates the circles (external division of the line of centres) and satisfies the condition of equal distance from both centres. The answer is $b+c = 12$.

   We have two circles. Let’s first rewrite them in standard form to find their centres and radii.

   Circle 1: $x^2 + y^2 - 2x - 10y + 1 = 0$

   Complete the square:

   $(x^2 - 2x) + (y^2 - 10y) = -1$

   $(x-1)^2 - 1 + (y-5)^2 - 25 = -1$

   $(x-1)^2 + (y-5)^2 = 25$

   So centre $C_1 = (1, 5)$, radius $r_1 = 5$.

   Circle 2: $x^2 + y^2 + 8x + 14y + 1 = 0$

   $(x^2 + 8x) + (y^2 + 14y) = -1$

   $(x+4)^2 - 16 + (y+7)^2 - 49 = -1$

   $(x+4)^2 + (y+7)^2 = 64$

   So centre $C_2 = (-4, -7)$, radius $r_2 = 8$.

   Now, a transverse common tangent is one that crosses the line joining the centres between the circles. For such a tangent, the distances from the two centres to the tangent line are in the ratio of the radii, but with the tangent lying between the circles — meaning the signed distances have opposite signs.

   Important

   For a transverse common tangent, if the line is $5x + by + c = 0$, then the perpendicular distances from $C_1$ and $C_2$ to this line satisfy:

   $\frac{|5(1) + b(5) + c|}{\sqrt{25 + b^2}} = r_1$ and $\frac{|5(-4) + b(-7) + c|}{\sqrt{25 + b^2}} = r_2$,

   but with opposite signs because the centres lie on opposite sides of the tangent.

   Let’s set up the signed distances. Let $d_1 = \frac{5(1) + 5b + c}{\sqrt{25+b^2}}$ and $d_2 = \frac{5(-4) -7b + c}{\sqrt{25+b^2}}$.

   For a transverse tangent, $d_1$ and $d_2$ have opposite signs, and $|d_1| = r_1 = 5$, $|d_2| = r_2 = 8$.

   So we have two cases: either $d_1 = 5$ and $d_2 = -8$, or $d_1 = -5$ and $d_2 = 8$. Both are symmetric; we’ll solve one.

   Take $d_1 = 5$, $d_2 = -8$. Then:

   1. $5 + 5b + c = 5\sqrt{25+b^2}$
   2. $-20 -7b + c = -8\sqrt{25+b^2}$

   From (1): $c = 5\sqrt{25+b^2} - 5 - 5b$

   From (2): $c = -8\sqrt{25+b^2} + 20 + 7b$ …

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The length of the common chord of the circles $x^2 + y^2 - 8x - 6y - 11 = 0$ and $x^2 + y^2 - 6x + 8y + 9 = 0$ is (A) $\sqrt{42}$ (B) $7$ (C) $\dfrac{13}{2}$ (D) $\sqrt{46}$

   ›Reveal solutionSolution

   The common chord is $x+7y+10=0$; the chord length from circle 1 is $2\sqrt{r^2-d^2}=\sqrt{46}$.

   Subtracting the two circle equations

   $(x^2+y^2-8x-6y-11)-(x^2+y^2-6x+8y+9)=0$

   gives $-2x-14y-20=0$, i.e. the common chord

   $x+7y+10=0.$

   Use the first circle $x^2+y^2-8x-6y-11=0$: centre $(4,3)$, radius $r=\sqrt{16+9+11}=6$.

   Distance from $(4,3)$ to the chord: …

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the length of a focal chord drawn to the parabola $y^2 = 64x$ is $289$ and the angle made by this focal chord with positive X-axis measured in positive direction is an acute angle then the slope of the focal chord is (A) $\dfrac{2}{5}$ (B) $\dfrac{5}{2}$ (C) $\dfrac{15}{8}$ (D) $\dfrac{8}{15}$

   ›Reveal solutionSolution

   For a parabola $y^2 = 4ax$, the length of a focal chord making an angle $\theta$ with the X-axis is $4a \csc^2 \theta$. Here $a = 16$, length $= 289$, so $\csc^2 \theta = \frac{289}{64}$, giving $\sin \theta = \frac{8}{17}$ and slope $\tan \theta = \frac{8}{15}$. The correct option is (D).

   The key idea is that a focal chord of a parabola passes through the focus, and its length can be expressed neatly in terms of the angle it makes with the axis. Once you know that formula, the problem becomes a straightforward trigonometric calculation.

   For the parabola $y^2 = 64x$, compare with the standard form $y^2 = 4ax$. Here $4a = 64$, so $a = 16$. The focus is at $(16, 0)$.

   1. Recall the formula for the length of a focal chord. For the parabola $y^2 = 4ax$, consider a focal chord making an angle $\theta$ with the positive X-axis (measured anticlockwise). The endpoints of this chord are the points where the line through the focus with slope $\tan \theta$ meets the parabola. The length of such a focal chord is given by:

   $L = 4a \csc^2 \theta$

   This is a standard result — it comes from solving the line $y = \tan \theta (x - a)$ with the parabola and using the distance formula.

   2. Plug in the given values. Here $a = 16$ and $L = 289$. So:

   $289 = 4 \times 16 \times \csc^2 \theta = 64 \csc^2 \theta$

   Therefore:

   $\csc^2 \theta = \frac{289}{64}$

   3. Find $\sin \theta$. Since $\csc \theta = \frac{1}{\sin \theta}$, we have:

   $\sin^2 \theta = \frac{64}{289}$

   Taking the positive square root (because $\theta$ is acute, so $\sin \theta > 0$):

   $\sin \theta = \frac{8}{17}$ …

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8. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Let $\dfrac{3}{2}$ be the eccentricity of the conjugate hyperbola of a hyperbola H and one of the foci of H lie on the straight line $x+y-3=0$. If the transverse and conjugate axes of H are along X, Y-axes respectively then the length of the latus rectum of H is (A) $\dfrac{4}{\sqrt{5}}$ (B) $\dfrac{8}{\sqrt{5}}$ (C) $\dfrac{10}{3}$ (D) $\dfrac{5}{3}$

   ›Reveal solutionSolution

   The eccentricity of the conjugate hyperbola gives the relationship between $a$ and $b$ for the original hyperbola $H$; using the focus condition on the line $x+y-3=0$ determines $a$ and $b$, and then the latus rectum length is computed as $\frac{8}{\sqrt{5}}$, which corresponds to option (B).

   ---

   Concept and Intuition

   We have a hyperbola $H$ with transverse axis along the X-axis and conjugate axis along the Y-axis. Its standard equation is

   $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

   with eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}}$.

   The conjugate hyperbola of $H$ is

   $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{or} \quad \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$

   Its eccentricity is $e' = \sqrt{1 + \frac{a^2}{b^2}}$.

   The problem gives $e' = \frac{3}{2}$. That directly links $a$ and $b$.

   Also, one focus of $H$ lies on the line $x+y-3=0$. For $H$, foci are at $(\pm ae, 0)$. So the condition becomes $|ae + 0 - 3| = 0$ (since the focus lies exactly on the line), giving $ae = 3$.

   From these two equations we can solve for $a$ and $b$, then find the latus rectum length $= \frac{2b^2}{a}$.

   ---

   Step-by-step solution

   1. Relate $a$ and $b$ using the conjugate hyperbola’s eccentricity For the conjugate hyperbola:

   $e' = \sqrt{1 + \frac{a^2}{b^2}} = \frac{3}{2}$

   Square both sides:

   $1 + \frac{a^2}{b^2} = \frac{9}{4} \quad\Rightarrow\quad \frac{a^2}{b^2} = \frac{5}{4}$

   So

   $a^2 = \frac{5}{4} b^2 \quad\text{or}\quad b^2 = \frac{4}{5} a^2$

   2. Use the focus condition The eccentricity of $H$ is

   $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{5}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$

   One focus is at $(ae, 0)$. It lies on $x+y-3=0$, so

   $ae + 0 - 3 = 0 \quad\Rightarrow\quad ae = 3$

   Substitute $e = \frac{3}{\sqrt{5}}$:

   $a \cdot \frac{3}{\sqrt{5}} = 3 \quad\Rightarrow\quad a = \sqrt{5}$ …

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9. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $(h, k)$ is the point to which the origin has to be shifted by translation of axes to remove the terms containing $x$ and $y$ from the equation $2x^2 + 3xy + y^2 + 4x - 8y + 5 = 0$ then $2h + k =$ (A) $0$ (B) $1$ (C) $20$ (D) $16$

   ›Reveal solutionSolution

   To eliminate the $x$ and $y$ terms from a general second-degree equation by translating the origin to $(h,k)$, we set the partial derivatives with respect to $x$ and $y$ to zero at $(h,k)$. Solving the resulting system of equations for $h$ and $k$ and then calculating $2h+k$ gives the result $\boxed{20}$.

   The problem asks us to find a specific point $(h, k)$ to which the origin must be shifted so that the linear terms (terms containing $x$ and $y$ individually) in the given equation vanish. This process is known as translation of axes.

   Concept and Intuition

   When we translate the coordinate axes, we are essentially moving the origin to a new point $(h, k)$ without rotating the axes. If a point had coordinates $(x, y)$ in the old system, its coordinates $(X, Y)$ in the new system will be related by:

   $x = X + h$

   $y = Y + k$

   Our goal is to choose $h$ and $k$ such that when we substitute these expressions into the original equation, the resulting equation in terms of $X$ and $Y$ does not contain any $X$ or $Y$ terms. This means the coefficients of $X$ and $Y$ must be zero.

   For a general second-degree equation $F(x,y) = ax^2 + 2h_{std}xy + by^2 + 2gx + 2fy + c = 0$, the coefficients of $X$ and $Y$ in the transformed equation (after substituting $x=X+h$ and $y=Y+k$) are directly related to the partial derivatives of $F(x,y)$ evaluated at the new origin $(h,k)$. Specifically, the coefficient of $X$ will be $\frac{\partial F}{\partial x}\Big|_{(h,k)}$ and the coefficient of $Y$ will be $\frac{\partial F}{\partial y}\Big|_{(h,k)}$ (up to a factor of 2, which doesn't affect the condition of being zero).

   Therefore, to eliminate the linear terms, we must satisfy the conditions:

   $\frac{\partial F}{\partial x}\Big|_{(h,k)} = 0$

   $\frac{\partial F}{\partial y}\Big|_{(h,k)} = 0$

   These two equations form a system of linear equations in $h$ and $k$, which we can solve to find the required translation point. This point $(h,k)$ is often referred to as the center of the conic section represented by the equation.

   Step-by-step Solution

   1. Identify coefficients from the given equation.

      The given equation is $2x^2 + 3xy + y^2 + 4x - 8y + 5 = 0$.

      We compare this with the general form $ax^2 + 2h_{std}xy + by^2 + 2gx + 2fy + c = 0$.

      From the comparison, we get:

      $a = 2$

      $2h_{std} = 3 \implies h_{std} = \frac{3}{2}$

      $b = 1$

      $2g = 4 \implies g = 2$

      $2f = -8 \implies f = -4$

      $c = 5$

   2. Formulate the system of equations for $(h,k)$.

      Let $F(x,y) = 2x^2 + 3xy + y^2 + 4x - 8y + 5$.

      First, calculate the partial derivatives:

      $\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3xy + y^2 + 4x - 8y + 5) = 4x + 3y + 4$ …

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10. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.Distance between the internal and external centres of similitude with respect to the circles $x^2 + y^2 + 4x + 6y + 12 = 0$ and $x^2 + y^2 - 6x - 4y + 9 = 0$ is (A) $5\sqrt{2}$ (B) $\dfrac{20\sqrt{2}}{3}$ (C) $16$ (D) $5$

    ›Reveal solutionSolution

    The internal and external centres of similitude divide the line of centres in the ratio $r_1:r_2$; the distance between them works out to $\dfrac{20\sqrt{2}}{3}$.

    Circle data.

    $C_1:x^2+y^2+4x+6y+12=0$ has centre $P_1=(-2,-3)$, radius $r_1=\sqrt{4+9-12}=1.$

    $C_2:x^2+y^2-6x-4y+9=0$ has centre $P_2=(3,2)$, radius $r_2=\sqrt{9+4-9}=2.$

    Distance between centres.

    $d=\sqrt{(3+2)^2+(2+3)^2}=\sqrt{25+25}=5\sqrt{2}.$

    Centres of similitude. Both lie on line $P_1P_2$. Measuring signed distance from $P_1$ toward $P_2$:

    Internal centre divides internally in $r_1:r_2=1:2$, at distance $\dfrac{r_1}{r_1+r_2}\,d=\dfrac{1}{3}d.$ …

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11. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.The distance of a point $(1,2)$ from the directrix of the parabola $y^2 - 4x - 4y + 8 = 0$ is (A) $2$ (B) $1$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    The key idea is to rewrite the given parabola in standard form to identify its directrix, then compute the perpendicular distance from the point $(1,2)$ to that line. The distance is $1$.

    The problem gives a second-degree equation that looks like a parabola but is not in the familiar $y^2 = 4ax$ form. The first step is always to complete the square to reveal the vertex and the axis. Once we have the standard form, the directrix is a straight line, and the distance from a point to a line is a straightforward formula.

    Let’s work through it.

    1. Rewrite the equation by completing the square in $y$. The given equation is

    $y^2 - 4x - 4y + 8 = 0.$

    Group the $y$ terms: $y^2 - 4y$. Complete the square:

    $y^2 - 4y = (y-2)^2 - 4.$

    Substitute back:

    $(y-2)^2 - 4 - 4x + 8 = 0 \quad\Rightarrow\quad (y-2)^2 - 4x + 4 = 0.$

    So

    $(y-2)^2 = 4x - 4 = 4(x-1).$

    2. Identify the standard form and parameters.

       The equation $(y-2)^2 = 4(x-1)$ matches the standard parabola $(y-k)^2 = 4a(x-h)$, where the vertex is at $(h,k) = (1,2)$ and $4a = 4$, so $a = 1$.

       This parabola opens to the right (since the $y$ term is squared and the $x$ term is positive).

    3. Find the directrix.

       For a right-opening parabola $(y-k)^2 = 4a(x-h)$, the directrix is the vertical line $x = h - a$.

       Here $h = 1$ and $a = 1$, so the directrix is

    $x = 1 - 1 = 0.$

    That is, the $y$-axis. …

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12. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.The locus of the midpoints of the chords of the circle $x^2 + y^2 - 2x - 2y + 1 = 0$ which are parallel to the line $x + y + 2 = 0$ is (A) $x - y = 2$ (B) $2x - 3y = 4$ (C) $3x + 4y = 2$ (D) $x - y = 0$

    ›Reveal solutionSolution

    The midpoints of all chords parallel to a given line lie on the line through the circle’s centre perpendicular to that direction. For this circle, the centre is (1,1) and the perpendicular slope is 1, so the locus is $x - y = 0$, which is option (D).

    Concept & Intuition

    When a set of parallel chords are drawn in a circle, their midpoints all lie on a straight line through the centre of the circle. This line is perpendicular to the direction of the chords. Why? Because the radius to the midpoint of a chord is perpendicular to the chord itself. So if all chords are parallel, their midpoints must lie on the line through the centre that is perpendicular to them. This is a standard property: the locus of midpoints of parallel chords of a circle is a diameter perpendicular to the chords.

    ---

    1. Rewrite the circle equation in centre-radius form Given:

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

    Complete the square:

    $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$

    $(x-1)^2 + (y-1)^2 = 1$

    So the centre is $C(1,1)$ and radius $r = 1$.

    2. Find the slope of the given line

       The line is $x + y + 2 = 0$, i.e. $y = -x - 2$. Its slope is $m = -1$.

    3. Determine the slope of the line containing the midpoints

       The chords are parallel to this line, so they also have slope $-1$. The line through the centre perpendicular to them will have slope equal to the negative reciprocal:

    $m_{\perp} = 1$

    (since $-1 \times 1 = -1$).

    4. Write the equation of the locus The line through the centre $(1,1)$ with slope $1$ is: …

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