Locus Related Problem Questions in English

(C) Let the two circles be $S_1 = 0$ and $S_2 = 0$.
Let the point be $P(x, y)$.
The square of the length of the tangent from $P$ to $S_1 = 0$ is $t_1^2 = S_1$.
The square of the length of the tangent from $P$ to $S_2 = 0$ is $t_2^2 = S_2$.
Given that $t_1^2 - t_2^2 = k$,where $k$ is a constant.
Substituting the expressions,we get $S_1 - S_2 = k$.
Since $S_1$ and $S_2$ are second-degree equations in $x$ and $y$ of the form $x^2 + y^2 + 2g_ix + 2f_iy + c_i = 0$,the difference $S_1 - S_2$ results in a linear equation of the form $2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = k$.
This is a first-degree equation in $x$ and $y$,which represents a straight line parallel to the radical axis $S_1 - S_2 = 0$.

(B) Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. This circle cuts the two given circles orthogonally.
The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Applying this to our circle,we get:
$2gg_1 + 2ff_1 = c + c_1$ .... $(i)$
$2gg_2 + 2ff_2 = c + c_2$ .... $(ii)$
Subtracting $(ii)$ from $(i)$,we get:
$2g(g_1 - g_2) + 2f(f_1 - f_2) = c_1 - c_2$
The center of the circle is $(-g, -f)$. Let the center be $(x, y)$,so $g = -x$ and $f = -y$.
Substituting these into the equation:
$2(-x)(g_1 - g_2) + 2(-y)(f_1 - f_2) = c_1 - c_2$
$-2x(g_1 - g_2) - 2y(f_1 - f_2) = c_1 - c_2$
$2x(g_1 - g_2) + 2y(f_1 - f_2) + c_1 - c_2 = 0$
This is the equation of the radical axis of the two given circles.

(D) Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.
It cuts the circle $x^2 + y^2 - 20x + 4 = 0$ orthogonally.
The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$2g(-10) + 2f(0) = c + 4$,which gives $-20g = c + 4$ $(ii)$.
Circle $(i)$ touches the line $x = 2$,so the perpendicular distance from the centre $(-g, -f)$ to the line $x - 2 = 0$ equals the radius $\sqrt{g^2 + f^2 - c}$.
$|(-g) - 2| = \sqrt{g^2 + f^2 - c} \Rightarrow (g + 2)^2 = g^2 + f^2 - c$.
$g^2 + 4g + 4 = g^2 + f^2 - c \Rightarrow 4g + 4 = f^2 - c$ $(iii)$.
Adding $(ii)$ and $(iii)$,we get $-20g + 4g + 4 = c + 4 + f^2 - c$.
$-16g + 4 = f^2 + 4 \Rightarrow f^2 = -16g$.
Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.
Substituting these,$(-y)^2 = -16(-x) \Rightarrow y^2 = 16x$.

(C) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since this circle cuts the circle $x^2 + y^2 + 4x - 6y + 9 = 0$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$:
$2g(2) + 2f(-3) = c + 9 \Rightarrow 4g - 6f = c + 9$ .....$(i)$
Since it also cuts $x^2 + y^2 - 4x + 6y + 4 = 0$ orthogonally:
$2g(-2) + 2f(3) = c + 4 \Rightarrow -4g + 6f = c + 4$ .....$(ii)$
Adding $(i)$ and $(ii)$ gives $0 = 2c + 13$,so $c = -\frac{13}{2}$.
Subtracting $(ii)$ from $(i)$ gives $8g - 12f = 5$.
The centre of the circle is $(-g, -f)$. Let the locus be $(x, y)$,so $x = -g$ and $y = -f$,which means $g = -x$ and $f = -y$.
Substituting these into the equation $8g - 12f = 5$ gives $8(-x) - 12(-y) = 5$,which simplifies to $-8x + 12y = 5$ or $8x - 12y + 5 = 0$.

(B) Let the point $P$ be $(x, y)$. The square of the length of the tangent from $P(x, y)$ to a circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $S_1 = x^2 + y^2 + 2gx + 2fy + c$.
Given the ratio of the squares of the lengths of the tangents is $2 : 3$:
$\frac{x^2 + y^2 + 2x - 4y - 20}{x^2 + y^2 - 4x + 2y - 44} = \frac{2}{3}$
$3(x^2 + y^2 + 2x - 4y - 20) = 2(x^2 + y^2 - 4x + 2y - 44)$
$3x^2 + 3y^2 + 6x - 12y - 60 = 2x^2 + 2y^2 - 8x + 4y - 88$
$x^2 + y^2 + 14x - 16y + 28 = 0$
Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = 14 \Rightarrow g = 7$ and $2f = -16 \Rightarrow f = -8$.
The centre of the circle is $(-g, -f) = (-7, 8)$.

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $x$-axis,the radius $r = |k|$.
The circle also touches the circle with centre $(0, 3)$ and radius $2$.
The distance between the centres is equal to the sum of the radii: $\sqrt{(h - 0)^2 + (k - 3)^2} = r + 2$.
Substituting $r = |k|$,we get $\sqrt{h^2 + (k - 3)^2} = |k| + 2$.
Squaring both sides: $h^2 + (k - 3)^2 = (|k| + 2)^2$.
$h^2 + k^2 - 6k + 9 = k^2 + 4|k| + 4$.
$h^2 = 4|k| + 6k + 5$.
If $k > 0$,$h^2 = 10k + 5 = 10(k + 0.5)$,which represents a parabola.
Thus,the locus of the centre $(x, y)$ is $x^2 = 10y + 5$.

(B) Given equations are $x = \frac{2at}{1 + t^2}$ and $y = \frac{a(1 - t^2)}{1 + t^2}$.
Squaring both equations,we get $x^2 = \frac{4a^2t^2}{(1 + t^2)^2}$ and $y^2 = \frac{a^2(1 - t^2)^2}{(1 + t^2)^2}$.
Adding $x^2$ and $y^2$,we have $x^2 + y^2 = \frac{4a^2t^2 + a^2(1 - 2t^2 + t^4)}{(1 + t^2)^2}$.
$x^2 + y^2 = \frac{a^2(4t^2 + 1 - 2t^2 + t^4)}{(1 + t^2)^2} = \frac{a^2(t^4 + 2t^2 + 1)}{(1 + t^2)^2}$.
Since $t^4 + 2t^2 + 1 = (1 + t^2)^2$,we get $x^2 + y^2 = \frac{a^2(1 + t^2)^2}{(1 + t^2)^2} = a^2$.
This represents a circle with radius $a$ centered at the origin.

(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.
Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.
Substituting the coordinates,we get $\frac{(x - a)^2 + y^2}{(x + a)^2 + y^2} = k^2$.
$(x - a)^2 + y^2 = k^2((x + a)^2 + y^2)$.
$(x^2 - 2ax + a^2 + y^2) = k^2(x^2 + 2ax + a^2 + y^2)$.
$(x^2 + y^2)(1 - k^2) - 2ax(1 + k^2) + a^2(1 - k^2) = 0$.
If $k = 1$,the equation becomes $-4ax = 0$,which represents a straight line (the perpendicular bisector of $AB$),not a circle.
Therefore,for the locus to be a circle,$k$ must not be equal to $1$.

(B) Let the variable point on the circle be $B(x_1, y_1)$. Since $B$ lies on the circle ${x^2} + {y^2} = 4$,we have ${x_1}^2 + {y_1}^2 = 4$.
Let the point dividing the segment $AB$ in the ratio $3 : 2$ be $P(h, k)$.
Using the section formula,we have:
$h = \frac{3x_1 + 2(-1)}{3 + 2} = \frac{3x_1 - 2}{5} \implies x_1 = \frac{5h + 2}{3}$
$k = \frac{3y_1 + 2(1)}{3 + 2} = \frac{3y_1 + 2}{5} \implies y_1 = \frac{5k - 2}{3}$
Substituting these values into the circle equation ${x_1}^2 + {y_1}^2 = 4$:
$(\frac{5h + 2}{3})^2 + (\frac{5k - 2}{3})^2 = 4$
$\frac{25h^2 + 20h + 4}{9} + \frac{25k^2 - 20k + 4}{9} = 4$
$25(h^2 + k^2) + 20(h - k) + 8 = 36$
$25(h^2 + k^2) + 20(h - k) - 28 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $25(x^2 + y^2) + 20(x - y) - 28 = 0$.

(D) The equation of the circle is ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$.
Comparing with ${x^2} + {y^2} + 2gx + 2fy + c = 0$,we have $g = 2, f = -3, c = 9{\sin ^2}\alpha + 13{\cos ^2}\alpha$.
The centre is $C(-g, -f) = (-2, 3)$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - (9{\sin ^2}\alpha + 13{\cos ^2}\alpha)} = \sqrt{13 - 9{\sin ^2}\alpha - 13{\cos ^2}\alpha} = \sqrt{13(1 - {\cos ^2}\alpha) - 9{\sin ^2}\alpha} = \sqrt{13{\sin ^2}\alpha - 9{\sin ^2}\alpha} = \sqrt{4{\sin ^2}\alpha} = 2\sin \alpha$.
Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2\alpha$,so $\angle APC = \alpha$.
In the right-angled triangle $\triangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC} = \frac{2\sin \alpha}{\sqrt{(h + 2)^2 + (k - 3)^2}}$.
Thus,$\sqrt{(h + 2)^2 + (k - 3)^2} = 2$.
Squaring both sides,$(h + 2)^2 + (k - 3)^2 = 4$.
Expanding,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$,which simplifies to $h^2 + k^2 + 4h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} + {y^2} + 4x - 6y + 9 = 0$.

(B) Let the coordinates of point $M$ be $(h, k)$.
Given $A = (0, 3)$ and $M = (h, k)$. Since $M$ lies on the line segment $AB$ such that $AM = 2AB$,point $B$ is the midpoint of $AM$.
Thus,the coordinates of $B$ are $\left( \frac{h+0}{2}, \frac{k+3}{2} \right) = \left( \frac{h}{2}, \frac{k+3}{2} \right)$.
Since $B$ lies on the circle $x^2 + 4x + (y - 3)^2 = 0$,we substitute the coordinates of $B$ into the circle equation:
$\left( \frac{h}{2} \right)^2 + 4\left( \frac{h}{2} \right) + \left( \frac{k+3}{2} - 3 \right)^2 = 0$
$\Rightarrow \frac{h^2}{4} + 2h + \left( \frac{k-3}{2} \right)^2 = 0$
$\Rightarrow \frac{h^2}{4} + 2h + \frac{(k-3)^2}{4} = 0$
Multiplying by $4$,we get $h^2 + 8h + (k-3)^2 = 0$,which simplifies to $h^2 + k^2 + 8h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 8x - 6y + 9 = 0$,which represents a circle.

(A) Let $(x', y')$ be the midpoint of a chord.
The equation of the chord with midpoint $(x', y')$ for the circle $x^2 + y^2 - 2x = 0$ is given by $T = S_1$.
$x x' + y y' - (x + x') = x'^2 + y'^2 - 2x'$
Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$0 + 0 - (0 + x') = x'^2 + y'^2 - 2x'$
$-x' = x'^2 + y'^2 - 2x'$
$x'^2 + y'^2 - x' = 0$
Replacing $(x', y')$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$,which represents a circle.

(B) Let the equation of the variable circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$ $... (i)$.
Since circle $(i)$ cuts the circle ${x^2} + {y^2} - 4 = 0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2g(0) + 2f(0) = c - 4$,which implies $c = 4$.
Since the circle $(i)$ passes through the point $(a, b)$,we have ${a^2} + {b^2} + 2ga + 2fb + 4 = 0$.
To find the locus of the centre $(-g, -f)$,let $x = -g$ and $y = -f$,so $g = -x$ and $f = -y$.
Substituting these into the equation,we get ${a^2} + {b^2} + 2(-x)a + 2(-y)b + 4 = 0$.
Thus,the locus is $2ax + 2by - ({a^2} + {b^2} + 4) = 0$.

(C) The equation of any tangent to the circle ${x^2} + {y^2} = {a^2}$ with slope $m$ is $y = mx \pm a\sqrt{1 + {m^2}}$.
If this tangent passes through the point $P(h, k)$,then $(k - mh)^2 = {a^2}(1 + {m^2})$.
Expanding this,we get ${k^2} - 2mhk + {m^2}{h^2} = {a^2} + {a^2}{m^2}$,which rearranges to ${m^2}({h^2} - {a^2}) - 2mhk + ({k^2} - {a^2}) = 0$.
Since $m = \tan \theta$,the roots of this quadratic equation in $m$ are ${m_1} = \tan {\theta _1}$ and ${m_2} = \tan {\theta _2}$.
Given $\cot {\theta _1} + \cot {\theta _2} = c$,we have $\frac{1}{{{m_1}}} + \frac{1}{{{m_2}}} = c$,which implies $\frac{{{m_1} + {m_2}}}{{{m_1}{m_2}}} = c$.
From the quadratic equation,the sum of roots ${m_1} + {m_2} = \frac{2hk}{{{h^2} - {a^2}}}$ and the product of roots ${m_1}{m_2} = \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}$.
Substituting these into the expression,we get $\frac{2hk}{{{k^2} - {a^2}}} = c$,which simplifies to $c({k^2} - {a^2}) = 2hk$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $c({y^2} - {a^2}) = 2xy$.

(A) Let the moving point be $P(h, k)$.
The distance of $P(h, k)$ from the point $(a, 0)$ is $\sqrt{(h - a)^2 + (k - 0)^2}$.
The distance of $P(h, k)$ from the $y$-axis (the line $x = 0$) is $|h|$.
According to the problem,these distances are equal:
$\sqrt{(h - a)^2 + k^2} = |h|$
Squaring both sides,we get:
$(h - a)^2 + k^2 = h^2$
$h^2 - 2ah + a^2 + k^2 = h^2$
$-2ah + a^2 + k^2 = 0$
Replacing $(h, k)$ with $(x, y)$,the equation of the locus is:
$y^2 - 2ax + a^2 = 0$.

(C) Let the two fixed straight lines be the coordinate axes $x$ and $y$. Let the bar have length $L = a + b$,where $a$ and $b$ are the segments of the bar divided by the point $P(h, k)$.
When the bar makes an angle $\theta$ with the $x$-axis,the coordinates of point $P$ are given by $h = a \cos \theta$ and $k = b \sin \theta$.
From these,we have $\cos \theta = \frac{h}{a}$ and $\sin \theta = \frac{k}{b}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$\frac{h^2}{a^2} + \frac{k^2}{b^2} = 1$.
This is the equation of an ellipse.

(C) Let the two given circles have centres $C_1$ and $C_2$ and radii $r_1$ and $r_2$ respectively.
Let the circle with centre $P(x, y)$ and radius $r$ touch these two circles externally.
Then,the distance $PC_1 = r + r_1$ and $PC_2 = r + r_2$.
Subtracting these equations,we get $PC_1 - PC_2 = (r + r_1) - (r + r_2) = r_1 - r_2$.
Since $r_1$ and $r_2$ are constants,$PC_1 - PC_2$ is a constant.
The locus of a point such that the difference of its distances from two fixed points (foci) is constant is a hyperbola.

(C) If an equation $f(x, y) = 0$ remains unchanged when $x$ is replaced by $-x$ and $y$ is replaced by $-y$,then $f(-x, -y) = f(x, y) = 0$.
This condition implies that for every point $(x, y)$ on the curve,the point $(-x, -y)$ also lies on the curve.
This geometric property is known as symmetry about the origin or symmetry in opposite quadrants.

(B) The given equation of the circle is $(x - 1)^2 + y^2 = 1$,which simplifies to $x^2 + y^2 - 2x = 0$.
Let the midpoint of the chord be $(h, k)$.
The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$.
Here,$T = xh + yk - (x + h)$ and $S_1 = h^2 + k^2 - 2h$.
So,the equation of the chord is $xh + yk - x - h = h^2 + k^2 - 2h$.
Since this chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$0(h) + 0(k) - 0 - h = h^2 + k^2 - 2h$.
$-h = h^2 + k^2 - 2h$.
$h^2 + k^2 = h$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = x$.

(D) Let $S = (x, y)$. The given relation is $SQ^2 + SR^2 = 2SP^2$.
Using the distance formula,we have:
$((x + 1)^2 + y^2) + ((x - 2)^2 + y^2) = 2((x - 1)^2 + y^2)$
Expanding the terms:
$(x^2 + 2x + 1 + y^2) + (x^2 - 4x + 4 + y^2) = 2(x^2 - 2x + 1 + y^2)$
$2x^2 - 2x + 5 + 2y^2 = 2x^2 - 4x + 2 + 2y^2$
Subtracting $2x^2 + 2y^2$ from both sides:
$-2x + 5 = -4x + 2$
$2x = -3$
$x = -\frac{3}{2}$
This represents a straight line parallel to the $y$-axis.

(D) Let the center of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $X$-axis,$r = |k|$.
The distance between the centers $(h, k)$ and $(0, 3)$ is $\sqrt{h^2 + (k - 3)^2}$.
Since the circles touch each other externally,the distance between their centers is equal to the sum of their radii: $\sqrt{h^2 + (k - 3)^2} = |k| + 2$.
Squaring both sides: $h^2 + (k - 3)^2 = (|k| + 2)^2$.
$h^2 + k^2 - 6k + 9 = k^2 + 4|k| + 4$.
$h^2 = 4|k| + 6k + 5 - 9 = 4|k| + 6k - 4$.
If $k > 0$,$h^2 = 4k + 6k - 4 = 10k - 4 = 10(k - 0.4)$.
This represents a parabola.

(A) The locus of the point of intersection of perpendicular tangents to a circle is called the director circle.
For a circle $x^2 + y^2 = r^2$,the equation of the director circle is $x^2 + y^2 = 2r^2$.
Here,the given circle is $x^2 + y^2 = 8$,so $r^2 = 8$.
The equation of the director circle is $x^2 + y^2 = 2(8) = 16$.
We are given that the point lies on the line $x = 3$.
Substituting $x = 3$ into the equation of the director circle:
$(3)^2 + y^2 = 16$
$9 + y^2 = 16$
$y^2 = 7$
$y = \pm \sqrt{7}$.
Thus,the points are $(3, \sqrt{7})$ and $(3, -\sqrt{7})$.
Comparing this with the given options,option $A$ is correct.

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it passes through $(1, 2)$,we have $1^2 + 2^2 + 2g(1) + 2f(2) + c = 0$,which simplifies to $2g + 4f + c + 5 = 0$.
Since the circle intersects $x^2 + y^2 = 4$ (or $x^2 + y^2 - 4 = 0$) orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ applies.
Here,$g_1 = g, f_1 = f, c_1 = c$ and $g_2 = 0, f_2 = 0, c_2 = -4$.
Thus,$2g(0) + 2f(0) = c - 4$,which gives $c = 4$.
Substituting $c = 4$ into the first equation: $2g + 4f + 4 + 5 = 0$,so $2g + 4f + 9 = 0$.
The center of the circle is $(-g, -f)$. Let the center be $(x, y)$,so $g = -x$ and $f = -y$.
Substituting these into the equation $2g + 4f + 9 = 0$,we get $2(-x) + 4(-y) + 9 = 0$,which simplifies to $2x + 4y - 9 = 0$.

(C) Let the coordinates of point $P$ be $(x, y)$.
Given $A = (c, 0)$ and $B = (-c, 0)$.
$PA^{2} = (x - c)^{2} + (y - 0)^{2} = x^{2} - 2cx + c^{2} + y^{2}$.
$PB^{2} = (x + c)^{2} + (y - 0)^{2} = x^{2} + 2cx + c^{2} + y^{2}$.
$AB^{2} = (c - (-c))^{2} + (0 - 0)^{2} = (2c)^{2} = 4c^{2}$.
Given the condition $PA^{2} + PB^{2} = AB^{2}$,we substitute the expressions:
$(x^{2} - 2cx + c^{2} + y^{2}) + (x^{2} + 2cx + c^{2} + y^{2}) = 4c^{2}$.
$2x^{2} + 2y^{2} + 2c^{2} = 4c^{2}$.
$2x^{2} + 2y^{2} = 2c^{2}$.
Dividing by $2$,we get $x^{2} + y^{2} = c^{2}$.

(B) The angle subtended by the chord at the center of the circle $x^{2} + y^{2} = 4$ is $2 \times 45^{\circ} = 90^{\circ}$.
Let the midpoint of the chord $AB$ be $(h, k)$. The equation of the chord $AB$ is given by $T = S_1$,which is:
$hx + ky - 4 = h^{2} + k^{2} - 4$
$hx + ky = h^{2} + k^{2}$
$\Rightarrow \frac{hx + ky}{h^{2} + k^{2}} = 1$
Using this line to homogenize the circle equation $x^{2} + y^{2} - 4 = 0$,we get:
$x^{2} + y^{2} - 4 \left( \frac{hx + ky}{h^{2} + k^{2}} \right)^{2} = 0$
This represents the combined equation of lines $OA$ and $OB$. Since the angle at the center is $90^{\circ}$,the lines $OA$ and $OB$ are perpendicular.
For perpendicular lines,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:
$(1 - \frac{4h^{2}}{(h^{2} + k^{2})^{2}}) + (1 - \frac{4k^{2}}{(h^{2} + k^{2})^{2}}) = 0$
$2 - \frac{4(h^{2} + k^{2})}{(h^{2} + k^{2})^{2}} = 0$
$2 - \frac{4}{h^{2} + k^{2}} = 0$
$h^{2} + k^{2} = 2$
Thus,the locus is $x^{2} + y^{2} = 2$.

(A) Let the circle passing through the ends of the rods be $x^2 + y^2 + 2gx + 2fy + c = 0$.
The rod of length $a$ lies on the $x$-axis,so its intercepts on the $x$-axis are the roots of $x^2 + 2gx + c = 0$. The length of the intercept is $2\sqrt{g^2 - c} = a$,which implies $4(g^2 - c) = a^2$.
The rod of length $b$ lies on the $y$-axis,so its intercepts on the $y$-axis are the roots of $y^2 + 2fy + c = 0$. The length of the intercept is $2\sqrt{f^2 - c} = b$,which implies $4(f^2 - c) = b^2$.
Subtracting the two equations: $4(g^2 - f^2) = a^2 - b^2$.
Since the center of the circle is $(h, k) = (-g, -f)$,we have $g = -h$ and $f = -k$.
Substituting these into the equation: $4((-h)^2 - (-k)^2) = a^2 - b^2$,which simplifies to $4(h^2 - k^2) = a^2 - b^2$.
Thus,the locus of the center $(h, k)$ is $4(x^2 - y^2) = a^2 - b^2$.

(A) Let the point be $P(x, y)$.
According to the problem,the distance of $P$ from $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.
$\sqrt{(x - 0)^2 + (y + 1)^2} = 2 \times \frac{|3x + 4y + 1|}{\sqrt{3^2 + 4^2}}$
$\sqrt{x^2 + (y + 1)^2} = 2 \times \frac{|3x + 4y + 1|}{5}$
$5\sqrt{x^2 + y^2 + 2y + 1} = 2|3x + 4y + 1|$
Squaring both sides:
$25(x^2 + y^2 + 2y + 1) = 4(3x + 4y + 1)^2$
$25(x^2 + y^2 + 2y + 1) = 4(9x^2 + 16y^2 + 1 + 24xy + 6x + 8y)$
$25x^2 + 25y^2 + 50y + 25 = 36x^2 + 64y^2 + 4 + 96xy + 24x + 32y$
Rearranging the terms to one side:
$(36 - 25)x^2 + (64 - 25)y^2 + 96xy + 24x + (32 - 50)y + (4 - 25) = 0$
$11x^2 + 39y^2 + 96xy + 24x - 18y - 21 = 0$

(A) Let $P(h, k)$ be the variable point and $A(-5, 1)$ and $B(3, 2)$ be the given points.
According to the given condition,$\angle APB = 90^{\circ}$.
This implies that the point $P$ lies on a circle with diameter $AB$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the coordinates of $A$ and $B$:
$(x - (-5))(x - 3) + (y - 1)(y - 2) = 0$
$(x + 5)(x - 3) + (y - 1)(y - 2) = 0$
$x^{2} + 2x - 15 + y^{2} - 3y + 2 = 0$
$x^{2} + y^{2} + 2x - 3y - 13 = 0$
Thus,the locus of $P(h, k)$ is $x^{2} + y^{2} + 2x - 3y - 13 = 0$.

(A) Let the other end of the diameter be $B(\alpha, \beta)$.
The equation of the circle with diameter $AB$ is:
$(x - p)(x - \alpha) + (y - q)(y - \beta) = 0$
$x^{2} + y^{2} - (p + \alpha)x - (q + \beta)y + p\alpha + q\beta = 0$ ... $(i)$
Since the circle touches the $x$-axis,the $y$-coordinate of the center is equal to the radius,or the discriminant of the equation obtained by setting $y = 0$ is zero.
Setting $y = 0$ in $(i)$:
$x^{2} - (p + \alpha)x + (p\alpha + q\beta) = 0$
Since it touches the $x$-axis,the discriminant $D = 0$:
$(p + \alpha)^{2} - 4(p\alpha + q\beta) = 0$
$p^{2} + 2p\alpha + \alpha^{2} - 4p\alpha - 4q\beta = 0$
$p^{2} - 2p\alpha + \alpha^{2} = 4q\beta$
$(p - \alpha)^{2} = 4q\beta$
Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $(x - p)^{2} = 4qy$.

(D) Let the coordinates of point $P$ be $(x, y)$.
Given the condition $2PA = 3PB$,we square both sides to get $4PA^{2} = 9PB^{2}$.
Using the distance formula,$PA^{2} = x^{2} + y^{2}$ and $PB^{2} = (x - 4)^{2} + (y + 3)^{2}$.
Substituting these into the equation: $4(x^{2} + y^{2}) = 9[(x - 4)^{2} + (y + 3)^{2}]$.
Expanding the terms: $4x^{2} + 4y^{2} = 9[x^{2} - 8x + 16 + y^{2} + 6y + 9]$.
$4x^{2} + 4y^{2} = 9x^{2} - 72x + 144 + 9y^{2} + 54y + 81$.
Rearranging the terms to one side: $5x^{2} + 5y^{2} - 72x + 54y + 225 = 0$.
Thus,the locus of point $P$ is $5x^{2} + 5y^{2} - 72x + 54y + 225 = 0$.

(D) Let the center of the required circle be $C_1 = (h, k)$. Since it touches the $y$-axis,its radius $r_1 = |h|$.
For the given circle $x^{2} + y^{2} - 6x - 6y + 14 = 0$,the center $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^{2} + 3^{2} - 14} = \sqrt{9 + 9 - 14} = \sqrt{4} = 2$.
Since the circles touch externally,the distance between their centers is $C_1 C_2 = r_1 + r_2$.
Thus,$\sqrt{(h - 3)^{2} + (k - 3)^{2}} = |h| + 2$.
Squaring both sides,$(h - 3)^{2} + (k - 3)^{2} = (|h| + 2)^{2}$.
$h^{2} - 6h + 9 + k^{2} - 6k + 9 = h^{2} + 4|h| + 4$.
$k^{2} - 6k + 14 = 6h + 4|h|$.
Assuming $h > 0$,$k^{2} - 6k + 14 = 10h$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2} - 6y + 14 = 10x$,or $y^{2} - 10x - 6y + 14 = 0$.

(B) Let the equation of the variable circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$ ... $(i)$.
Since the circle $(i)$ intersects the circle $x^{2} + y^{2} - 4 = 0$ orthogonally,the condition $2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$ applies.
Here,$g_{2} = 0, f_{2} = 0, c_{2} = -4$.
So,$2g(0) + 2f(0) = c - 4$,which gives $c = 4$.
Since the circle passes through the point $(a, b)$,we have $a^{2} + b^{2} + 2ga + 2fb + 4 = 0$.
To find the locus of the center $(-g, -f)$,let $x = -g$ and $y = -f$,so $g = -x$ and $f = -y$.
Substituting these into the equation: $a^{2} + b^{2} + 2(-x)a + 2(-y)b + 4 = 0$.
This simplifies to $a^{2} + b^{2} - 2ax - 2by + 4 = 0$,or $2ax + 2by - (a^{2} + b^{2} + 4) = 0$.

(A) Let the midpoint of the chord $AB$ of the circle $x^2 + y^2 = a^2$ be $P(h, k)$.
The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$,which is $hx + ky = h^2 + k^2$.
The combined equation of the lines $OA$ and $OB$ is obtained by homogenizing the circle equation $x^2 + y^2 = a^2$ using the chord equation:
$x^2 + y^2 = a^2 \left( \frac{hx + ky}{h^2 + k^2} \right)^2$
$(h^2 + k^2)(x^2 + y^2) = a^2(hx + ky)^2$
$(h^2 + k^2)(x^2 + y^2) - a^2(h^2x^2 + k^2y^2 + 2hkxy) = 0$
$(h^2 + k^2 - a^2h^2)x^2 + (h^2 + k^2 - a^2k^2)y^2 - 2a^2hkxy = 0$
Since the chord subtends a right angle at the center,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(h^2 + k^2 - a^2h^2) + (h^2 + k^2 - a^2k^2) = 0$
$2(h^2 + k^2) - a^2(h^2 + k^2) = 0$
$(h^2 + k^2)(2 - a^2) = 0$
Wait,re-evaluating: The distance from the center $(0,0)$ to the chord $hx + ky = h^2 + k^2$ is $d = \frac{|h^2 + k^2|}{\sqrt{h^2 + k^2}} = \sqrt{h^2 + k^2}$.
In a right-angled triangle formed by the center and the chord,the distance $d$ from the center to the chord is $a \cos(45^\circ) = \frac{a}{\sqrt{2}}$.
So,$\sqrt{h^2 + k^2} = \frac{a}{\sqrt{2}}$.
Squaring both sides,$h^2 + k^2 = \frac{a^2}{2}$,which gives $2(h^2 + k^2) - a^2 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2(x^2 + y^2) - a^2 = 0$.

(C) Let $AB$ be a chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center $C(0, 0)$.
Let $D$ be the midpoint of the chord $AB$. Then $CD \perp AB$.
In $\Delta ACD$,the angle $\angle ACD = \frac{1}{2} \angle ACB = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}$.
The radius $CA = 3$.
Let the coordinates of the midpoint $D$ be $(h, k)$. Then $CD = \sqrt{h^2 + k^2}$.
In the right-angled triangle $\Delta ACD$,we have $\cos(\angle ACD) = \frac{CD}{CA}$.
$\cos(\frac{\pi}{3}) = \frac{\sqrt{h^2 + k^2}}{3}$.
$\frac{1}{2} = \frac{\sqrt{h^2 + k^2}}{3}$.
$\sqrt{h^2 + k^2} = \frac{3}{2}$.
Squaring both sides,we get $h^2 + k^2 = \frac{9}{4}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = \frac{9}{4}$.

(C) The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. The equation of a tangent to this hyperbola with slope $m$ is $y = mx \pm \sqrt{16m^2 - 9} \dots (i)$.
Let $(x_1, y_1)$ be the midpoint of the chord of the circle $x^2 + y^2 = 16$. The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$,which is $xx_1 + yy_1 = x_1^2 + y_1^2 \dots (ii)$.
Comparing $(i)$ and $(ii)$,we rewrite $(ii)$ as $y = -\frac{x_1}{y_1}x + \frac{x_1^2 + y_1^2}{y_1}$.
Equating the slopes and intercepts: $m = -\frac{x_1}{y_1}$ and $\sqrt{16m^2 - 9} = \frac{x_1^2 + y_1^2}{y_1}$.
Squaring both sides: $16m^2 - 9 = \frac{(x_1^2 + y_1^2)^2}{y_1^2}$.
Substituting $m = -\frac{x_1}{y_1}$: $16\left(-\frac{x_1}{y_1}\right)^2 - 9 = \frac{(x_1^2 + y_1^2)^2}{y_1^2}$.
$16x_1^2 - 9y_1^2 = (x_1^2 + y_1^2)^2$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 = 16x^2 - 9y^2$.

(D) The circle is $x^2 + y^2 + 4x - 6y + 9\sin^2\alpha + 13\cos^2\alpha = 0$.
Its center is $C(-2, 3)$.
The radius $r$ is given by $\sqrt{(-2)^2 + (3)^2 - (9\sin^2\alpha + 13\cos^2\alpha)} = \sqrt{4 + 9 - 9\sin^2\alpha - 13\cos^2\alpha} = \sqrt{13 - 9\sin^2\alpha - 13\cos^2\alpha} = \sqrt{13(1 - \cos^2\alpha) - 9\sin^2\alpha} = \sqrt{13\sin^2\alpha - 9\sin^2\alpha} = \sqrt{4\sin^2\alpha} = 2\sin\alpha$.
Let $P(h, k)$ be the point. The angle between the tangents is $2\alpha$,so the angle between the line $PC$ and a tangent is $\alpha$.
In the right-angled triangle $\triangle PAC$,$\sin\alpha = \frac{AC}{PC} = \frac{r}{PC}$.
Thus,$\sin\alpha = \frac{2\sin\alpha}{\sqrt{(h+2)^2 + (k-3)^2}}$.
This implies $\sqrt{(h+2)^2 + (k-3)^2} = 2$.
Squaring both sides,$(h+2)^2 + (k-3)^2 = 4$.
Expanding this,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$.
$h^2 + k^2 + 4h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 4x - 6y + 9 = 0$.

(A) Let the third vertex be $P(x, y)$.
Since the triangle is a right-angled triangle with the hypotenuse connecting $(2, 0)$ and $(0, 2)$,the angle at vertex $P$ is $90^{\circ}$.
The product of the slopes of the two sides meeting at $P$ must be $-1$.
Slope $m_1 = \frac{y - 0}{x - 2} = \frac{y}{x - 2}$.
Slope $m_2 = \frac{y - 2}{x - 0} = \frac{y - 2}{x}$.
Setting $m_1 \times m_2 = -1$:
$\frac{y}{x - 2} \times \frac{y - 2}{x} = -1$
$y(y - 2) = -x(x - 2)$
$y^2 - 2y = -x^2 + 2x$
$x^2 + y^2 - 2x - 2y = 0$.

(D) Let the point be $P(x, y)$.
The distance of point $P(x, y)$ from the $X$-axis is $|y|$.
The distance of point $P(x, y)$ from the $Y$-axis is $|x|$.
According to the problem,the sum of the squares of these distances is $4$.
Therefore,$|x|^2 + |y|^2 = 4$.
This simplifies to $x^2 + y^2 = 4$.

(D) Let the point be $(h, k)$.
Given the ratio of distances is $\frac{1}{3}$:
$\frac{\sqrt{(h-1)^2 + k^2}}{\sqrt{(h+1)^2 + k^2}} = \frac{1}{3}$
Squaring both sides:
$9((h-1)^2 + k^2) = (h+1)^2 + k^2$
$9(h^2 - 2h + 1 + k^2) = h^2 + 2h + 1 + k^2$
$9h^2 - 18h + 9 + 9k^2 = h^2 + 2h + 1 + k^2$
$8h^2 + 8k^2 - 20h + 8 = 0$
Dividing by $8$:
$h^2 + k^2 - \frac{5}{2}h + 1 = 0$
Completing the square for $h$:
$(h^2 - \frac{5}{2}h + \frac{25}{16}) + k^2 = \frac{25}{16} - 1$
$(h - \frac{5}{4})^2 + k^2 = \frac{9}{16} = (\frac{3}{4})^2$
This is the equation of a circle with center $(\frac{5}{4}, 0)$ and radius $\frac{3}{4}$.
Since points $A, B$,and $C$ lie on this circle,the triangle $ABC$ is inscribed in this circle.
The circumcenter of any triangle inscribed in a circle is the center of that circle.
Therefore,the circumcenter is $(\frac{5}{4}, 0)$.

(B) Given the coordinates of point $P$ are $x = \cos \theta + \sin \theta$ and $y = \sin \theta - \cos \theta$.
To find the locus,we eliminate the parameter $\theta$ by squaring and adding the expressions for $x$ and $y$:
$x^2 + y^2 = (\cos \theta + \sin \theta)^2 + (\sin \theta - \cos \theta)^2$
Expanding the squares:
$x^2 + y^2 = (\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta) + (\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta)$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$x^2 + y^2 = (1 + 2 \sin \theta \cos \theta) + (1 - 2 \sin \theta \cos \theta)$
Simplifying the expression:
$x^2 + y^2 = 1 + 1 = 2$
Thus,the locus of $P$ is $x^2 + y^2 = 2$.

(C) The given equation is of the form $PF_1 + PF_2 = 2a$,where $P = (x, y)$,$F_1 = (2, -1)$,and $F_2 = (-2, -4)$.
First,calculate the distance between the two fixed points $F_1$ and $F_2$:
$d = \sqrt{(2 - (-2))^2 + (-1 - (-4))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Here,the sum of the distances from point $P$ to two fixed points $F_1$ and $F_2$ is equal to the distance between the two fixed points $(PF_1 + PF_2 = F_1F_2 = 5)$.
When the sum of distances from a point to two fixed points equals the distance between the fixed points,the locus of the point $P$ is the line segment joining the two fixed points $F_1$ and $F_2$.
Therefore,the equation represents a line segment.

(B) Let $P = (x, y)$. The condition $PA = nPB$ implies $PA^2 = n^2 PB^2$.
$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.
$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2ax + a^2 + y^2)$.
$(n^2 - 1)x^2 + (n^2 - 1)y^2 + 2ax(n^2 + 1) + a^2(n^2 - 1) = 0$.
Dividing by $(n^2 - 1)$,we get $x^2 + y^2 + 2ax \frac{n^2 + 1}{n^2 - 1} + a^2 = 0$.
This is the equation of a circle with center $C = (-a \frac{n^2 + 1}{n^2 - 1}, 0)$ and radius $r = |\frac{2an}{n^2 - 1}|$.
Since $a < 0$,let $a = -k$ where $k > 0$. Then the center is $C = (k \frac{n^2 + 1}{n^2 - 1}, 0)$.
For $n > 1$,the circle encloses the point $B(-a, 0)$ because $B$ is closer to the center relative to the radius,while $A(a, 0)$ lies outside.

(A) Let the point $P$ be $(h, k)$.
The equation of the pair of tangents from $P(h, k)$ to the circle $x^2 + y^2 = a^2$ is given by $SS_1 = T^2$.
$(x^2 + y^2 - a^2)(h^2 + k^2 - a^2) = (xh + yk - a^2)^2$.
Expanding this,the coefficient of $x^2$ is $(h^2 + k^2 - a^2) - h^2 = k^2 - a^2$.
The coefficient of $y^2$ is $(h^2 + k^2 - a^2) - k^2 = h^2 - a^2$.
Since the tangents are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
$(k^2 - a^2) + (h^2 - a^2) = 0 \implies h^2 + k^2 = 2a^2$.
However,the problem states the tangents to the first circle are perpendicular to the tangents to the second circle. This implies the locus of $P$ is the director circle of the system. For the condition where tangents from $P$ to $x^2 + y^2 = a^2$ are perpendicular to tangents from $P$ to $x^2 + y^2 = b^2$,the locus is $x^2 + y^2 = a^2 + b^2$.

(A) Let the midpoint of the chord of contact be $(h, k)$.
The equation of the chord of contact with midpoint $(h, k)$ for the circle $x^2 + y^2 = 9$ is given by $hx + ky = h^2 + k^2$ ... $(1)$.
Let a point on the line $4x - 5y = 20$ be $P(\alpha, \beta)$,where $4\alpha - 5\beta = 20$,so $\beta = \frac{4\alpha - 20}{5}$.
The equation of the chord of contact from point $P(\alpha, \beta)$ to the circle $x^2 + y^2 = 9$ is $\alpha x + \beta y = 9$ ... $(2)$.
Comparing $(1)$ and $(2)$,we have $\frac{h}{\alpha} = \frac{k}{\beta} = \frac{h^2 + k^2}{9}$.
Thus,$\alpha = \frac{9h}{h^2 + k^2}$ and $\beta = \frac{9k}{h^2 + k^2}$.
Substitute these into the line equation $4\alpha - 5\beta = 20$:
$4\left(\frac{9h}{h^2 + k^2}\right) - 5\left(\frac{9k}{h^2 + k^2}\right) = 20$.
$36h - 45k = 20(h^2 + k^2)$.
Replacing $(h, k)$ with $(x, y)$,the locus is $20(x^2 + y^2) - 36x + 45y = 0$.

(B) Let the points be $A(5, 0)$ and $B(10 \cos \theta, 10 \sin \theta)$.
Using the section formula,the coordinates of point $P(x, y)$ dividing $AB$ in the ratio $2 : 3$ are:
$x = \frac{2(10 \cos \theta) + 3(5)}{2 + 3} = \frac{20 \cos \theta + 15}{5} = 4 \cos \theta + 3$
$y = \frac{2(10 \sin \theta) + 3(0)}{2 + 3} = \frac{20 \sin \theta}{5} = 4 \sin \theta$
From these equations,we have:
$x - 3 = 4 \cos \theta \implies \cos \theta = \frac{x - 3}{4}$
$y = 4 \sin \theta \implies \sin \theta = \frac{y}{4}$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$(\frac{x - 3}{4})^2 + (\frac{y}{4})^2 = 1$
$(x - 3)^2 + y^2 = 16$
This is the equation of a circle with center $(3, 0)$ and radius $4$.

(A) Let the point $P$ be $(h, k)$.
Since the tangents are perpendicular,the angle between them is $90^{\circ}$.
The locus of the intersection of two perpendicular tangents to a circle is known as the director circle.
For a circle $x^2 + y^2 = a^2$,the radius is $a$ and the center is $(0, 0)$.
The distance from the center $(0, 0)$ to the point $P(h, k)$ is $\sqrt{h^2 + k^2}$.
In the right-angled triangle formed by the center,the point of contact,and $P$,the distance $PC$ is given by $\sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$.
Thus,$\sqrt{h^2 + k^2} = a\sqrt{2}$.
Squaring both sides,we get $h^2 + k^2 = 2a^2$.
Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $x^2 + y^2 = 2a^2$.

(B) Let the coordinates of point $P$ be $(x, y)$.
Given $PA = nPB$,squaring both sides gives $PA^2 = n^2 PB^2$.
Substituting the coordinates,we get $(x - a)^2 + (y - 0)^2 = n^2 ((x + a)^2 + (y - 0)^2)$.
Expanding the terms: $x^2 - 2ax + a^2 + y^2 = n^2 (x^2 + 2ax + a^2 + y^2)$.
Rearranging the terms: $x^2(1 - n^2) + y^2(1 - n^2) - 2ax(1 + n^2) + a^2(1 - n^2) = 0$.
Since $|n| \neq 1$,we have $1 - n^2 \neq 0$. Dividing by $(1 - n^2)$,we get $x^2 + y^2 - 2ax \frac{1 + n^2}{1 - n^2} + a^2 = 0$.
This is the equation of a circle of the form $x^2 + y^2 + 2gx + 2fy + c = 0$.

(A) The given circle is $x^2 + y^2 + 4x - 2y - 4 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2, f = -1, c = -4$.
The center is $C(-g, -f) = (-2, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 - (-4)} = \sqrt{9} = 3$.
Let $P(h, k)$ be a point on the locus. The angle between the tangents is $60^{\circ}$,so the angle between the line joining the center to $P$ and the radius is $30^{\circ}$.
In the right-angled triangle formed by the center,the point of contact,and $P$,we have $\sin(30^{\circ}) = \frac{r}{CP}$.
$\frac{1}{2} = \frac{3}{CP} \implies CP = 6$.
$CP^2 = 36$.
$(h + 2)^2 + (k - 1)^2 = 36$.
$h^2 + 4h + 4 + k^2 - 2k + 1 = 36$.
$h^2 + k^2 + 4h - 2k - 31 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 4x - 2y - 31 = 0$.