Locus Related Problem Questions in English

(C) Let the new coordinates be $(x', y')$ such that $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + y^2 - 4x + 6y - 7 = 0$ gives:
$(x' + h)^2 + (y' + k)^2 - 4(x' + h) + 6(y' + k) - 7 = 0$
Expanding the terms:
$x'^2 + 2hx' + h^2 + y'^2 + 2ky' + k^2 - 4x' - 4h + 6y' + 6k - 7 = 0$
Grouping the terms:
$x'^2 + y'^2 + x'(2h - 4) + y'(2k + 6) + (h^2 + k^2 - 4h + 6k - 7) = 0$
To eliminate the linear terms,the coefficients of $x'$ and $y'$ must be zero:
$2h - 4 = 0 \Rightarrow h = 2$
$2k + 6 = 0 \Rightarrow k = -3$
Thus,the point $(h, k)$ is $(2, -3)$.

(B) Let the point be $P(h, k)$.
Given that the distance of $P$ from $(a, 0)$ is equal to its distance from the $y$-axis.
The distance of $P(h, k)$ from $(a, 0)$ is $\sqrt{(h-a)^2 + (k-0)^2}$.
The distance of $P(h, k)$ from the $y$-axis is $|h|$.
Equating the distances: $\sqrt{(h-a)^2 + k^2} = |h|$.
Squaring both sides: $(h-a)^2 + k^2 = h^2$.
$h^2 - 2ah + a^2 + k^2 = h^2$.
$k^2 - 2ah + a^2 = 0$.
Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $y^2 - 2ax + a^2 = 0$.

(B) To find the locus,we eliminate the parameter $\theta$.
Given equations are:
$x = a(1 - \cos \theta )$ $\Rightarrow x - a = -a \cos \theta$ $\Rightarrow \cos \theta = -\frac{x - a}{a} = \frac{a - x}{a}$ .....$(i)$
$y = a \sin \theta \Rightarrow \sin \theta = \frac{y}{a}$ .....$(ii)$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we substitute $(i)$ and $(ii)$:
$(\frac{y}{a})^2 + (\frac{a - x}{a})^2 = 1$
$\frac{y^2}{a^2} + \frac{a^2 - 2ax + x^2}{a^2} = 1$
$y^2 + a^2 - 2ax + x^2 = a^2$
$x^2 + y^2 - 2ax = 0$
This is the equation of a circle with center $(a, 0)$ and radius $a$.

(D) Let the coordinates of point $S$ be $(x, y)$.
Given the relation $SQ^2 + SR^2 = 2SP^2$.
Substituting the coordinates,we get:
$((x + 1)^2 + (y - 0)^2) + ((x - 2)^2 + (y - 0)^2) = 2((x - 1)^2 + (y - 0)^2)$
$(x^2 + 2x + 1 + y^2) + (x^2 - 4x + 4 + y^2) = 2(x^2 - 2x + 1 + y^2)$
$2x^2 - 2x + 5 + 2y^2 = 2x^2 - 4x + 2 + 2y^2$
$-2x + 5 = -4x + 2$
$2x = -3$
$x = -\frac{3}{2}$
This represents a straight line parallel to the $y$-axis.

(D) Let the point be $P(x, y)$.
Given that $PA^2 + PB^2 = AB^2$.
The distance $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$,so $AB^2 = 16$.
The distance $PA^2 = (x - 2)^2 + (y - 0)^2 = x^2 - 4x + 4 + y^2$.
The distance $PB^2 = (x + 2)^2 + (y - 0)^2 = x^2 + 4x + 4 + y^2$.
Summing these: $(x^2 - 4x + 4 + y^2) + (x^2 + 4x + 4 + y^2) = 16$.
$2x^2 + 2y^2 + 8 = 16$.
$2x^2 + 2y^2 = 8$.
$x^2 + y^2 = 4$,which is $x^2 + y^2 - 4 = 0$.

(B) Let the moving 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 origin $(0, 0)$ is $\sqrt{x^2 + y^2}$.
According to the given condition,the distance from the $x$-axis is one-half of its distance from the origin:
$|y| = \frac{1}{2} \sqrt{x^2 + y^2}$
Squaring both sides,we get:
$y^2 = \frac{1}{4} (x^2 + y^2)$
$4y^2 = x^2 + y^2$
$x^2 + y^2 - 4y^2 = 0$
$x^2 - 3y^2 = 0$
Thus,the required equation of the locus is $x^2 - 3y^2 = 0$.

(B) Let the point be $P(x, y)$.
According to the given condition,the distance from $P(x, y)$ to $(-1, 0)$ is $3$ times the distance from $P(x, y)$ to $(0, 2)$.
$\sqrt{(x + 1)^2 + (y - 0)^2} = 3 \sqrt{(x - 0)^2 + (y - 2)^2}$
Squaring both sides:
$(x + 1)^2 + y^2 = 9(x^2 + (y - 2)^2)$
$x^2 + 2x + 1 + y^2 = 9(x^2 + y^2 - 4y + 4)$
$x^2 + 2x + 1 + y^2 = 9x^2 + 9y^2 - 36y + 36$
Rearranging the terms to one side:
$8x^2 + 8y^2 - 2x - 36y + 35 = 0$
This equation is of the form $Ax^2 + Ay^2 + Dx + Ey + F = 0$,which represents a circle.

(B) Let the coordinates of the point be $(x, y)$.
The distance of the point $(x, y)$ from the $x$-axis is $|y|$.
The distance of the point $(x, y)$ from the $y$-axis is $|x|$.
According to the given condition,the distance from the $x$-axis is double the distance from the $y$-axis:
$|y| = 2|x|$
Squaring both sides,we get $y^2 = 4x^2$,which represents two lines $y = 2x$ and $y = -2x$.
Among the given options,the equation $y = 2x$ represents the locus.

(D) Let the coordinates of $P$ be $(x, y)$.
Given $PA = kPB$,squaring both sides gives $PA^2 = k^2 PB^2$.
Using the distance formula: $(x - ak)^2 + y^2 = k^2 \left[ (x - \frac{a}{k})^2 + y^2 \right]$.
Expanding the terms: $x^2 - 2akx + a^2k^2 + y^2 = k^2 \left[ x^2 - 2x(\frac{a}{k}) + \frac{a^2}{k^2} + y^2 \right]$.
$x^2 - 2akx + a^2k^2 + y^2 = k^2x^2 - 2akx + a^2 + k^2y^2$.
Rearranging the terms: $x^2(1 - k^2) + y^2(1 - k^2) = a^2 - a^2k^2$.
$(1 - k^2)(x^2 + y^2) = a^2(1 - k^2)$.
Since $k = \pm 1$,$k^2 = 1$,so $1 - k^2 = 0$. However,for the locus to be defined as a standard equation independent of $k$ when $k \neq 1$,we simplify the expression: $x^2 + y^2 = a^2$.
Thus,the equation is $x^2 + y^2 - a^2 = 0$.

(A) Let the point be $(x, y)$.
The distance of $(x, y)$ from $(0, 0)$ is $\sqrt{x^2 + y^2}$.
The distance of $(x, y)$ from the $x$-axis is $|y|$.
According to the given condition,the distance from $(0, 0)$ is three times the distance from the $x$-axis:
$\sqrt{x^2 + y^2} = 3|y|$
Squaring both sides,we get:
$x^2 + y^2 = 9y^2$
$x^2 + y^2 - 9y^2 = 0$
$x^2 - 8y^2 = 0$

(A) Let the point be $P(x, y)$.
The distance between $P(x, y)$ and $(-g, -f)$ is given as $a$.
Using the distance formula: $\sqrt{(x - (-g))^2 + (y - (-f))^2} = a$
Squaring both sides: $(x + g)^2 + (y + f)^2 = a^2$
Expanding the squares: $x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = a^2$
Rearranging the terms: $x^2 + y^2 + 2gx + 2fy + g^2 + f^2 - a^2 = 0$
Given that $k = g^2 + f^2 - a^2$,we substitute $k$ into the equation:
$x^2 + y^2 + 2gx + 2fy + k = 0$.

(D) Let the coordinates of point $P$ be $(h, k)$.
Given the condition $2PA = 3PB$,we square both sides to get $4PA^2 = 9PB^2$.
Substituting the coordinates $A(0, 0)$ and $B(4, -3)$,we have:
$4(h^2 + k^2) = 9((h - 4)^2 + (k + 3)^2)$
$4h^2 + 4k^2 = 9(h^2 - 8h + 16 + k^2 + 6k + 9)$
$4h^2 + 4k^2 = 9h^2 - 72h + 144 + 9k^2 + 54k + 81$
Rearranging the terms to one side:
$5h^2 + 5k^2 - 72h + 54k + 225 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is:
$5x^2 + 5y^2 - 72x + 54y + 225 = 0$.

(C) Let the point be $P(x, y)$.
Given that the distance of $P$ from $A(1, -2)$ is twice the distance of $P$ from $B(-3, 5)$.
So,$PA = 2PB \Rightarrow PA^2 = 4PB^2$.
$(x - 1)^2 + (y + 2)^2 = 4[(x + 3)^2 + (y - 5)^2]$.
$x^2 - 2x + 1 + y^2 + 4y + 4 = 4[x^2 + 6x + 9 + y^2 - 10y + 25]$.
$x^2 + y^2 - 2x + 4y + 5 = 4[x^2 + y^2 + 6x - 10y + 34]$.
$x^2 + y^2 - 2x + 4y + 5 = 4x^2 + 4y^2 + 24x - 40y + 136$.
Rearranging the terms,we get:
$3x^2 + 3y^2 + 26x - 44y + 131 = 0$.

(B) Let the coordinates of the moving point be $(x, y)$.
Given that the distance of the point from the origin $(0, 0)$ is always $4$.
Using the distance formula,we have $\sqrt{(x - 0)^2 + (y - 0)^2} = 4$.
Squaring both sides,we get $x^2 + y^2 = 4^2$.
Therefore,the locus of the point is $x^2 + y^2 = 16$.

(D) Let the point be $P(x, y)$. Since $\angle APB = 90^\circ$,the slopes of $AP$ and $BP$ are perpendicular.
Slope of $AP = \frac{y - 0}{x - (-a)} = \frac{y}{x + a}$
Slope of $BP = \frac{y - 0}{x - a} = \frac{y}{x - a}$
Since the product of slopes is $-1$:
$\left(\frac{y}{x + a}\right) \times \left(\frac{y}{x - a}\right) = -1$
$\frac{y^2}{x^2 - a^2} = -1$
$y^2 = -(x^2 - a^2)$
$y^2 = -x^2 + a^2$
$x^2 + y^2 = a^2$

(B) Let the coordinates of $A$ be $(-a, 0)$ and $B$ be $(a, 0)$.
Let the coordinates of $P$ be $(x, y)$.
Given $PA^2 + PB^2 = k$.
$(x + a)^2 + y^2 + (x - a)^2 + y^2 = k$.
$(x^2 + 2ax + a^2 + y^2) + (x^2 - 2ax + a^2 + y^2) = k$.
$2x^2 + 2y^2 + 2a^2 = k$.
$x^2 + y^2 = \frac{k - 2a^2}{2}$.
This is the equation of a circle.

(B) Let $(h, k)$ be the centroid of the triangle. The coordinates of the centroid are given by the average of the vertices:
$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$
Rearranging the terms,we get:
$3h - 1 = \cos \alpha + \sin \alpha$ and $3k - 2 = \sin \alpha - \cos \alpha$
Squaring both equations and adding them:
$(3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2$
$(9h^2 - 6h + 1) + (9k^2 - 12k + 4) = (\cos^2 \alpha + \sin^2 \alpha + 2\sin \alpha \cos \alpha) + (\sin^2 \alpha + \cos^2 \alpha - 2\sin \alpha \cos \alpha)$
$9h^2 + 9k^2 - 6h - 12k + 5 = 1 + 1 = 2$
$9h^2 + 9k^2 - 6h - 12k + 3 = 0$
Dividing by $3$,we get:
$3(h^2 + k^2) - 2h - 4k + 1 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $3(x^2 + y^2) - 2x - 4y + 1 = 0$.

(B) Let the point be $P(x, y)$.
The distance of $P$ from the origin $(0, 0)$ is $OP = \sqrt{x^2 + y^2}$.
The distance of $P$ from the line $x = 2$ is $|x - 2|$.
Given that the sum of these distances is $4$,we have $\sqrt{x^2 + y^2} + |x - 2| = 4$.
Assuming $x \leq 2$,the equation becomes $\sqrt{x^2 + y^2} = 4 - (2 - x) = 2 + x$ (if $x < 2$) or $\sqrt{x^2 + y^2} = 4 - (x - 2) = 6 - x$ (if $x \geq 2$).
Considering the standard form,we take $\sqrt{x^2 + y^2} = 6 - x$.
Squaring both sides,we get $x^2 + y^2 = (6 - x)^2$.
$x^2 + y^2 = 36 + x^2 - 12x$.
$y^2 = 36 - 12x$.
$y^2 + 12x = 36$.

(B) Let the coordinates of the moving point be $(x, y)$.
The distance of the point $(x, y)$ from the $x$-axis is $|y|$.
The distance of the point $(x, y)$ from the origin $(0, 0)$ is $\sqrt{x^2 + y^2}$.
According to the problem,$4 \times (\text{distance from } x\text{-axis}) = (\text{distance from origin})^2$.
Substituting the values,we get $4|y| = (\sqrt{x^2 + y^2})^2$.
This simplifies to $4|y| = x^2 + y^2$.
Rearranging the terms,we get $x^2 + y^2 - 4|y| = 0$.

(A) Let the point be $P(x, y)$.
Given that the square of the distance from $P(x, y)$ to $(3, -2)$ is equal to the distance from $P(x, y)$ to the line $5x - 12y - 13 = 0$.
$(x - 3)^2 + (y + 2)^2 = \frac{|5x - 12y - 13|}{\sqrt{5^2 + (-12)^2}}$
$(x^2 - 6x + 9) + (y^2 + 4y + 4) = \frac{|5x - 12y - 13|}{13}$
$13(x^2 + y^2 - 6x + 4y + 13) = |5x - 12y - 13|$
$13x^2 + 13y^2 - 78x + 52y + 169 = \pm(5x - 12y - 13)$
Case $1$: $13x^2 + 13y^2 - 78x + 52y + 169 = 5x - 12y - 13$
$13x^2 + 13y^2 - 83x + 64y + 182 = 0$
Case $2$: $13x^2 + 13y^2 - 78x + 52y + 169 = -5x + 12y + 13$
$13x^2 + 13y^2 - 73x + 40y + 156 = 0$
Comparing with the given options,the equation $13x^2 + 13y^2 - 83x + 64y + 182 = 0$ matches option $A$.

(A) Let the point be $P(h, k)$.
According to the given condition,the distance of $P$ from $(4, 0)$ is half of its distance from the line $x - 16 = 0$.
So,$\sqrt{(h - 4)^2 + (k - 0)^2} = \frac{1}{2} |h - 16|$.
Squaring both sides,we get $(h - 4)^2 + k^2 = \frac{1}{4} (h - 16)^2$.
$4(h^2 - 8h + 16 + k^2) = h^2 - 32h + 256$.
$4h^2 - 32h + 64 + 4k^2 = h^2 - 32h + 256$.
$3h^2 + 4k^2 = 192$.
Replacing $(h, k)$ with $(x, y)$,the locus is $3x^2 + 4y^2 = 192$.

(B) Let the coordinates of the point $P(h, k)$ divide the line segment joining $(1, 0)$ and $(2\cos \theta, 2\sin \theta)$ in the ratio $2 : 3$.
Using the section formula,we have:
$h = \frac{2(2\cos \theta) + 3(1)}{2 + 3} = \frac{4\cos \theta + 3}{5}$
$k = \frac{2(2\sin \theta) + 3(0)}{2 + 3} = \frac{4\sin \theta}{5}$
From these,we get:
$\cos \theta = \frac{5h - 3}{4}$ and $\sin \theta = \frac{5k}{4}$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute the values:
$\left(\frac{5h - 3}{4}\right)^2 + \left(\frac{5k}{4}\right)^2 = 1$
$(5h - 3)^2 + (5k)^2 = 16$
Replacing $(h, k)$ with $(x, y)$,the locus is $(5x - 3)^2 + (5y)^2 = 16$,which represents a circle.

(B) Given the lines:
$1) x \cos \theta + y \sin \theta = a$
$2) x \sin \theta - y \cos \theta = b$
Squaring and adding both equations:
$(x \cos \theta + y \sin \theta)^2 + (x \sin \theta - y \cos \theta)^2 = a^2 + b^2$
$x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta + x^2 \sin^2 \theta + y^2 \cos^2 \theta - 2xy \sin \theta \cos \theta = a^2 + b^2$
$x^2(\cos^2 \theta + \sin^2 \theta) + y^2(\sin^2 \theta + \cos^2 \theta) = a^2 + b^2$
$x^2 + y^2 = a^2 + b^2$
This represents the equation of a circle with center $(0, 0)$ and radius $\sqrt{a^2 + b^2}$.

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre of the circle is $(-g, -f)$.
The length of the intercept cut off by the circle on the $x$-axis is $2\sqrt{g^2 - c} = 2a$,which implies $g^2 - c = a^2$ $(i)$.
The length of the intercept cut off by the circle on the $y$-axis is $2\sqrt{f^2 - c} = 2b$,which implies $f^2 - c = b^2$ $(ii)$.
Subtracting $(ii)$ from $(i)$,we get $g^2 - f^2 = a^2 - b^2$.
Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.
Substituting these into the equation,we get $(-x)^2 - (-y)^2 = a^2 - b^2$,which simplifies to $x^2 - y^2 = a^2 - b^2$.

(D) Let the center of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $y$-axis,the radius $r = |h|$.
Since the circle also touches the line $x - y = 0$,the perpendicular distance from the center $(h, k)$ to the line must be equal to the radius $r$.
Using the distance formula,$\frac{|h - k|}{\sqrt{1^2 + (-1)^2}} = |h|$.
This simplifies to $|h - k| = |h|\sqrt{2}$.
Squaring both sides,$(h - k)^2 = 2h^2$,which gives $h^2 - 2hk + k^2 = 2h^2$,or $h^2 + 2hk - k^2 = 0$.
For any given $k \neq 0$,this quadratic equation in $h$ provides real values for $h$.
Since there are infinitely many values for $k$,there are infinitely many such circles.

(C) Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
The length of the chord cut by the circle on the $x$-axis $(y=0)$ is $2\sqrt{g^2 - c} = 2a$,which implies $g^2 - c = a^2$,or $c = g^2 - a^2$ $(i)$.
The circle passes through the point $(0, b)$ on the $y$-axis,so $0^2 + b^2 + 2g(0) + 2f(b) + c = 0$,which gives $b^2 + 2fb + c = 0$ $(ii)$.
Substituting $c = g^2 - a^2$ from $(i)$ into $(ii)$,we get $b^2 + 2fb + g^2 - a^2 = 0$.
Rearranging,we have $g^2 + 2fb = a^2 - b^2$.
Replacing $(g, f)$ with $(x, y)$ to find the locus,we get $x^2 + 2by = a^2 - b^2$.

(D) Let the circle be $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.
Since the circle touches the $x$-axis,the radius $r = |k|$.
Thus,the equation of the circle is $(x - h)^2 + (y - k)^2 = k^2$,which simplifies to $x^2 - 2hx + h^2 + y^2 - 2ky = 0$.
The circle cuts a chord of length $2l$ from the $y$-axis (where $x = 0$).
Substituting $x = 0$ into the equation,we get $y^2 - 2ky + h^2 = 0$.
The length of the intercept on the $y$-axis is given by $2\sqrt{k^2 - h^2} = 2l$.
Squaring both sides,we get $k^2 - h^2 = l^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 - x^2 = l^2$,which represents a hyperbola.

(B) Let the two fixed coplanar points be $A(0, 0)$ and $B(a, 0)$.
Let the moving point $P$ be $(x, y)$.
According to the given condition,the ratio of distances is constant,say $\lambda$:
$\frac{PA}{PB} = \lambda \implies \frac{\sqrt{x^2 + y^2}}{\sqrt{(x-a)^2 + y^2}} = \lambda$
Squaring both sides:
$x^2 + y^2 = \lambda^2 ((x-a)^2 + y^2)$
$x^2 + y^2 = \lambda^2 (x^2 - 2ax + a^2 + y^2)$
$(1 - \lambda^2)x^2 + (1 - \lambda^2)y^2 + 2a\lambda^2 x - a^2\lambda^2 = 0$
Since $\lambda \ne 1$,we can divide by $(1 - \lambda^2)$:
$x^2 + y^2 + \frac{2a\lambda^2}{1 - \lambda^2}x - \frac{a^2\lambda^2}{1 - \lambda^2} = 0$
This is the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Thus,the locus is a circle.

(D) The given equation is $x^2 + y^2 + 4x + 6y + 13 = 0$.
Rearranging the terms,we get $(x^2 + 4x) + (y^2 + 6y) = -13$.
Completing the square for $x$ and $y$:
$(x^2 + 4x + 4) + (y^2 + 6y + 9) = -13 + 4 + 9$.
$(x + 2)^2 + (y + 3)^2 = 0$.
This is of the form $(x - h)^2 + (y - k)^2 = r^2$,where $r^2 = 0$.
Since the radius is $0$,the equation represents a point circle at $(-2, -3)$.

(C) Let the radius of the circle be $r$. Since the circle touches both coordinate axes,the distance from the centre $(h, k)$ to both axes must be equal to the radius.
Thus,$|h| = |k| = r$.
This implies $h = k$ or $h = -k$.
Therefore,the locus of the centre $(x, y)$ is given by $x = y$ or $x = -y$.
Combining these,we get $(x - y)(x + y) = 0$,which simplifies to $x^2 - y^2 = 0$.

(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Let the moving point be $P(x, y)$.
The sum of the squares of the distances from $P$ to the vertices is constant,say $K$:
$(x - x_1)^2 + (y - y_1)^2 + (x - x_2)^2 + (y - y_2)^2 + (x - x_3)^2 + (y - y_3)^2 = K$
Expanding the terms:
$3x^2 - 2x(x_1 + x_2 + x_3) + 3y^2 - 2y(y_1 + y_2 + y_3) + (x_1^2 + x_2^2 + x_3^2 + y_1^2 + y_2^2 + y_3^2) = K$
Dividing by $3$:
$x^2 - 2x\left(\frac{x_1 + x_2 + x_3}{3}\right) + y^2 - 2y\left(\frac{y_1 + y_2 + y_3}{3}\right) = \text{constant}$
This is the equation of a circle with centre at $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$,which is the centroid of the triangle.

(C) The locus of the point of intersection of perpendicular tangents to a circle is known as the director circle.
For a circle $x^2 + y^2 = a^2$,the director circle is given by $x^2 + y^2 = 2a^2$.
This represents a concentric circle with the original circle,having a radius of $\sqrt{2a^2} = a\sqrt{2}$.
Thus,the correct option is $C$.

(D) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $y$-axis,the radius $r = |h|$. Assuming $h > 0$,we have $r = h$.
The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. Its centre $C_1$ is $(3, 3)$ and its radius $R_1 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = \sqrt{4} = 2$.
Since the circles touch externally,the distance between their centres $C(h, k)$ and $C_1(3, 3)$ is equal to the sum of their radii:
$CC_1 = R_1 + r$
$\sqrt{(h - 3)^2 + (k - 3)^2} = 2 + h$
Squaring both sides:
$(h - 3)^2 + (k - 3)^2 = (2 + h)^2$
$h^2 - 6h + 9 + k^2 - 6k + 9 = 4 + 4h + h^2$
$k^2 - 6k + 18 - 6h = 4 + 4h$
$k^2 - 10h - 6k + 14 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 - 10x - 6y + 14 = 0$.

(B) Let $(h, k)$ be the centre of the circle of radius $r_2 = 2$ that rolls on the outside of the given circle.
The given circle is ${x^2} + {y^2} + 3x - 6y - 9 = 0$.
Its centre $C_1$ is $\left( -\frac{3}{2}, 3 \right)$ and its radius $r_1 = \sqrt{\left( \frac{3}{2} \right)^2 + (-3)^2 - (-9)} = \sqrt{\frac{9}{4} + 9 + 9} = \sqrt{\frac{9 + 72}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.
Since the circle rolls on the outside,the distance between the centres $(h, k)$ and $\left( -\frac{3}{2}, 3 \right)$ is equal to the sum of the radii,$r_1 + r_2 = \frac{9}{2} + 2 = \frac{13}{2}$.
Thus,the locus satisfies: $\left( h + \frac{3}{2} \right)^2 + (k - 3)^2 = \left( \frac{13}{2} \right)^2$.
Expanding this: $h^2 + 3h + \frac{9}{4} + k^2 - 6k + 9 = \frac{169}{4}$.
$h^2 + k^2 + 3h - 6k + \frac{9 + 36 - 169}{4} = 0$.
$h^2 + k^2 + 3h - 6k - \frac{124}{4} = 0$.
$h^2 + k^2 + 3h - 6k - 31 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} + {y^2} + 3x - 6y - 31 = 0$.

(D) Let the centre of the circle be $(h, k)$.
Since the circle touches the $y$-axis,the radius of the circle is equal to the absolute value of the $x$-coordinate of the centre,i.e.,$r = |h|$.
Since the circle passes through the point $(2, 0)$,the distance from the centre $(h, k)$ to $(2, 0)$ must equal the radius $r$.
Thus,$(h - 2)^2 + (k - 0)^2 = h^2$.
Expanding this,we get $h^2 - 4h + 4 + k^2 = h^2$.
Simplifying,we get $k^2 = 4h - 4$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $y^2 = 4x - 4$,which represents a parabola.

(B) Let the centre of the circle be $O(x, y)$.
Since the circle touches the $Y$-axis,the radius of the circle is $r = |x|$.
The perpendicular drawn from the centre $O(x, y)$ to the chord on the $X$-axis bisects the chord.
Let the chord be $AB$ of length $2a$. The midpoint of the chord is $M(x, 0)$.
The distance from the centre $O(x, y)$ to the midpoint $M(x, 0)$ is $|y|$.
In the right-angled triangle formed by the centre,the midpoint of the chord,and one endpoint of the chord,the hypotenuse is the radius $r = |x|$.
By Pythagoras theorem,$r^2 = |y|^2 + a^2$.
Substituting $r = x$,we get $x^2 = y^2 + a^2$,which simplifies to $x^2 - y^2 = a^2$.

(C) Let the centre of the circle be $(h, k)$.
Since the circle passes through the point $(a, 0)$,the radius $r$ is the distance between $(h, k)$ and $(a, 0)$,so $r = \sqrt{(h - a)^2 + k^2}$.
The circle also touches the line $x + 1 = 0$,so the radius $r$ is the perpendicular distance from $(h, k)$ to the line $x + 1 = 0$,which is $r = |h + 1|$.
Equating the two expressions for $r^2$,we get $(h - a)^2 + k^2 = (h + 1)^2$.
Expanding this,$h^2 - 2ah + a^2 + k^2 = h^2 + 2h + 1$.
Simplifying,$k^2 = 2h(1 + a) + 1 - a^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2x(1 + a) + (1 - a^2)$,which represents a parabola.

(B) The condition for the line $lx + my - 1 = 0$ to be a tangent to the circle ${x^2} + {y^2} = {a^2}$ is that the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $a$.
Applying the formula for the distance from a point to a line: $\frac{|l(0) + m(0) - 1|}{\sqrt{l^2 + m^2}} = a$.
This simplifies to $\frac{1}{\sqrt{l^2 + m^2}} = a$.
Squaring both sides,we get $l^2 + m^2 = \frac{1}{a^2}$.
Replacing $(l, m)$ with $(x, y)$,the locus is $x^2 + y^2 = \frac{1}{a^2}$,which represents a circle.

(C) The director circle of a given circle is the locus of the point of intersection of two perpendicular tangents to the circle.
For a circle with radius $r$,the director circle is a concentric circle with radius $r\sqrt{2}$.
Given the circle ${x^2} + {y^2} = {a^2}$,the radius is $r = a$.
Therefore,the radius of the director circle is $a\sqrt{2}$.
The equation of the director circle is ${x^2} + {y^2} = (a\sqrt{2})^2$,which simplifies to ${x^2} + {y^2} = 2{a^2}$.

(C) Let $P(x_1, y_1)$ be a point on the circle $x^2 + y^2 = 4$.
The equation of the tangent at $P$ is $xx_1 + yy_1 = 4$.
This tangent meets the coordinate axes at $A(\frac{4}{x_1}, 0)$ and $B(0, \frac{4}{y_1})$.
Let $(h, k)$ be the mid-point of $AB$.
Then $h = \frac{2}{x_1}$ and $k = \frac{2}{y_1}$,which implies $x_1 = \frac{2}{h}$ and $y_1 = \frac{2}{k}$.
Since $(x_1, y_1)$ lies on $x^2 + y^2 = 4$,we have $(\frac{2}{h})^2 + (\frac{2}{k})^2 = 4$.
Dividing by $4$,we get $\frac{1}{h^2} + \frac{1}{k^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = 1$.

(B) The condition for the line $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ to touch the circle $x^2 + y^2 = a^2$ is that the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $a$.
The equation of the line can be written as $\frac{1}{\alpha}x + \frac{1}{\beta}y - 1 = 0$.
The perpendicular distance $d$ is given by $d = \frac{|\frac{1}{\alpha}(0) + \frac{1}{\beta}(0) - 1|}{\sqrt{(\frac{1}{\alpha})^2 + (1/\beta)^2}} = a$.
Squaring both sides,we get $\frac{1}{(\frac{1}{\alpha})^2 + (1/\beta)^2} = a^2$.
This simplifies to $(\frac{1}{\alpha})^2 + (1/\beta)^2 = \frac{1}{a^2}$.
Let $X = \frac{1}{\alpha}$ and $Y = \frac{1}{\beta}$. The equation becomes $X^2 + Y^2 = (\frac{1}{a})^2$,which represents a circle.

(D) Let the point be $(x_1, y_1)$.
The length of the tangent from $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
According to the question,the ratio of the lengths of the tangents is $2:3$:
$\frac{\sqrt{x_1^2 + y_1^2 + 4x_1 + 3}}{\sqrt{x_1^2 + y_1^2 - 6x_1 + 5}} = \frac{2}{3}$
Squaring both sides,we get:
$\frac{x_1^2 + y_1^2 + 4x_1 + 3}{x_1^2 + y_1^2 - 6x_1 + 5} = \frac{4}{9}$
Cross-multiplying:
$9(x_1^2 + y_1^2 + 4x_1 + 3) = 4(x_1^2 + y_1^2 - 6x_1 + 5)$
$9x_1^2 + 9y_1^2 + 36x_1 + 27 = 4x_1^2 + 4y_1^2 - 24x_1 + 20$
Rearranging the terms:
$5x_1^2 + 5y_1^2 + 60x_1 + 7 = 0$
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $5x^2 + 5y^2 + 60x + 7 = 0$.

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

(D) Let the midpoint of the chord be $(h, k)$.
The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$,where $T$ is the tangent-like expression and $S_1$ is the value of the circle equation at $(h, k)$.
The equation of the circle is $S: x^2 + y^2 - 2x - 6y - 10 = 0$.
Thus,$hx + ky - (x + h) - 3(y + k) - 10 = h^2 + k^2 - 2h - 6k - 10$.
Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$h(0) + k(0) - (0 + h) - 3(0 + k) - 10 = h^2 + k^2 - 2h - 6k - 10$.
$-h - 3k - 10 = h^2 + k^2 - 2h - 6k - 10$.
Simplifying the equation:
$h^2 + k^2 - h - 3k = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - x - 3y = 0$.

(A) The given circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -1, c = -2$.
The centre $C$ is $(-g, -f) = (1, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 1 - (-2)} = \sqrt{4} = 2$.
Let $M(h, k)$ be the midpoint of a chord $AB$. The triangle $CAM$ is a right-angled triangle at $M$,where $CA = r = 2$ and $\angle ACM = \frac{120^\circ}{2} = 60^\circ = \frac{\pi}{3}$ radians.
In $\triangle CAM$,$\cos(\angle ACM) = \frac{CM}{CA}$.
$\cos(60^\circ) = \frac{CM}{2}$ $\Rightarrow \frac{1}{2} = \frac{CM}{2}$ $\Rightarrow CM = 1$.
The distance $CM$ is the distance between $(h, k)$ and $(1, 1)$,so $CM^2 = (h-1)^2 + (k-1)^2$.
Since $CM = 1$,we have $(h-1)^2 + (k-1)^2 = 1^2$.
$h^2 - 2h + 1 + k^2 - 2k + 1 = 1$.
$h^2 + k^2 - 2h - 2k + 1 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 2x - 2y + 1 = 0$.

(D) Let the line passing through the origin $(0,0)$ be $y = mx$,or $mx - y = 0$.
Let $(h, k)$ be the centre of the circle drawn with $AB$ as the diameter.
The centre of the circle $x^2 + y^2 - 2ax = 0$ is $C(a, 0)$.
Since the line $AB$ is a chord of the circle,the perpendicular from the centre $C(a, 0)$ to the chord $AB$ bisects the chord at its midpoint $M(h, k)$.
Thus,the line $CM$ is perpendicular to the line $AB$.
The slope of $AB$ is $m = \frac{k}{h}$.
The slope of $CM$ is $\frac{k - 0}{h - a} = \frac{k}{h - a}$.
Since $CM \perp AB$,the product of their slopes is $-1$:
$\left(\frac{k}{h}\right) \times \left(\frac{k}{h - a}\right) = -1$
$k^2 = -h(h - a)$
$k^2 = -h^2 + ah$
$h^2 + k^2 - ah = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - ax = 0$.

(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 cuts $x^2 + y^2 = 4$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ applies,where $g_1 = g, f_1 = f, c_1 = c$ and $g_2 = 0, f_2 = 0, c_2 = -4$.
This gives $2g(0) + 2f(0) = c - 4$,so $c = 4$.
Substituting $c = 4$ into the first equation: $2g + 4f + 4 + 5 = 0$,which gives $2g + 4f + 9 = 0$.
Replacing $g = -x$ and $f = -y$ for the locus of the centre $(-g, -f)$,we get $2(-x) + 4(-y) + 9 = 0$,or $2x + 4y - 9 = 0$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it cuts the circle $x^2 + y^2 = p^2$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2g(0) + 2f(0) = c - p^2$,which implies $c = p^2$.
Since the circle passes through $(a, b)$,we have $a^2 + b^2 + 2ga + 2fb + p^2 = 0$.
The centre of the circle is $(-g, -f)$. Let the centre be $(x, y)$,so $g = -x$ and $f = -y$.
Substituting these into the equation,we get $a^2 + b^2 + 2(-x)a + 2(-y)b + p^2 = 0$.
This simplifies to $2ax + 2by - (a^2 + b^2 + p^2) = 0$.