Locus Related Problem Questions in English

(B) Let $Q = (3 \cos \theta, 3 \sin \theta)$ be a point on the circle $x^2+y^2=9$.
Let $P = (h, k)$ be the point that divides the line segment $QA$ in the ratio $1:2$.
Using the section formula,the coordinates of $P$ are given by:
$h = \frac{1(4) + 2(3 \cos \theta)}{1+2} = \frac{4 + 6 \cos \theta}{3}$
$k = \frac{1(4) + 2(3 \sin \theta)}{1+2} = \frac{4 + 6 \sin \theta}{3}$
Rearranging these equations:
$3h - 4 = 6 \cos \theta$
$3k - 4 = 6 \sin \theta$
Squaring and adding both equations:
$(3h - 4)^2 + (3k - 4)^2 = (6 \cos \theta)^2 + (6 \sin \theta)^2$
$9(h - \frac{4}{3})^2 + 9(k - \frac{4}{3})^2 = 36$
$(h - \frac{4}{3})^2 + (k - \frac{4}{3})^2 = 4$
This represents a circle with radius $r = \sqrt{4} = 2$.
The perimeter of the locus is $2 \pi r = 2 \pi (2) = 4 \pi$.

(B) Let $(x, y)$ be the point which is equidistant from the point $(1, 1)$ and the line $x+y+1=0$.
By the definition of distance,the distance from $(x, y)$ to $(1, 1)$ is $\sqrt{(x-1)^2+(y-1)^2}$.
The distance from $(x, y)$ to the line $x+y+1=0$ is $\frac{|x+y+1|}{\sqrt{1^2+1^2}} = \frac{|x+y+1|}{\sqrt{2}}$.
Equating the two distances:
$\sqrt{(x-1)^2+(y-1)^2} = \frac{|x+y+1|}{\sqrt{2}}$
Squaring both sides:
$(x-1)^2+(y-1)^2 = \frac{(x+y+1)^2}{2}$
$2(x^2-2x+1+y^2-2y+1) = x^2+y^2+1+2xy+2x+2y$
$2x^2+2y^2-4x-4y+4 = x^2+y^2+2xy+2x+2y+1$
$x^2+y^2-2xy-6x-6y+3 = 0$
$(x-y)^2-6(x+y)+3 = 0$.

(A) Given $x = \frac{3at}{1+t^3}$ and $y = \frac{3at^2}{1+t^3}$.
We observe that $xy = \frac{9a^2t^3}{(1+t^3)^2}$ and $x^2y = \frac{27a^3t^4}{(1+t^3)^3}$,which is not the direct path.
Instead,consider $x^3 + y^3 = \left(\frac{3at}{1+t^3}\right)^3 + \left(\frac{3at^2}{1+t^3}\right)^3$.
$x^3 + y^3 = \frac{27a^3t^3 + 27a^3t^6}{(1+t^3)^3} = \frac{27a^3t^3(1+t^3)}{(1+t^3)^3} = \frac{27a^3t^3}{(1+t^3)^2}$.
Now,$3axy = 3a \left(\frac{3at}{1+t^3}\right) \left(\frac{3at^2}{1+t^3}\right) = \frac{27a^3t^3}{(1+t^3)^2}$.
Therefore,$x^3 + y^3 = 3axy$.

(A) Let the coordinates of point $P$ be $(x, y)$.
Given $PA + PB = 8$.
$\sqrt{(x-2)^2 + (y-3)^2} + \sqrt{(x-2)^2 + (y+3)^2} = 8$
$\sqrt{(x-2)^2 + (y-3)^2} = 8 - \sqrt{(x-2)^2 + (y+3)^2}$
Squaring both sides:
$(x-2)^2 + (y-3)^2 = 64 + (x-2)^2 + (y+3)^2 - 16\sqrt{(x-2)^2 + (y+3)^2}$
$y^2 - 6y + 9 = 64 + y^2 + 6y + 9 - 16\sqrt{(x-2)^2 + (y+3)^2}$
$-12y - 64 = -16\sqrt{(x-2)^2 + (y+3)^2}$
$3y + 16 = 4\sqrt{(x-2)^2 + (y+3)^2}$
Squaring again:
$9y^2 + 96y + 256 = 16((x-2)^2 + (y+3)^2)$
$9y^2 + 96y + 256 = 16(x^2 - 4x + 4 + y^2 + 6y + 9)$
$9y^2 + 96y + 256 = 16x^2 - 64x + 16y^2 + 96y + 208$
$16x^2 + 7y^2 - 64x - 48 = 0$.

(A) Let $A = (-1, 0)$ and $B = (0, 2)$.
Given the ratio of distances $\frac{PA}{PB} = \frac{\sqrt{2}}{1}$.
Squaring both sides,we get $\frac{PA^2}{PB^2} = 2$.
$PA^2 = (x+1)^2 + (y-0)^2 = x^2 + 2x + 1 + y^2$.
$PB^2 = (x-0)^2 + (y-2)^2 = x^2 + y^2 - 4y + 4$.
Substituting into the ratio equation: $x^2 + 2x + 1 + y^2 = 2(x^2 + y^2 - 4y + 4)$.
$x^2 + 2x + 1 + y^2 = 2x^2 + 2y^2 - 8y + 8$.
Rearranging terms: $x^2 + y^2 - 2x - 8y + 7 = 0$.
Completing the square: $(x^2 - 2x + 1) + (y^2 - 8y + 16) = -7 + 1 + 16$.
$(x-1)^2 + (y-4)^2 = 10$.

(B) Let the point be $P(x, y)$.
According to the given condition,the sum of the squares of the distances from $(a, 0)$ and $(-a, 0)$ is $2b^2$.
Using the distance formula,we have:
$((x-a)^2 + (y-0)^2) + ((x+a)^2 + (y-0)^2) = 2b^2$
Expanding the squares:
$(x^2 - 2ax + a^2 + y^2) + (x^2 + 2ax + a^2 + y^2) = 2b^2$
Combining like terms:
$2x^2 + 2y^2 + 2a^2 = 2b^2$
Dividing the entire equation by $2$:
$x^2 + y^2 + a^2 = b^2$
Rearranging the terms,we get the locus of $P$ as:
$x^2 + y^2 = b^2 - a^2$

(D) Let the point be $(x, y)$.
The distance between $(x, y)$ and $(3, -2)$ is given as $4$ units.
Using the distance formula,$\sqrt{(x-3)^2 + (y-(-2))^2} = 4$.
Squaring both sides,we get $(x-3)^2 + (y+2)^2 = 16$.
Expanding the terms,$x^2 - 6x + 9 + y^2 + 4y + 4 = 16$.
Simplifying the equation,$x^2 + y^2 - 6x + 4y + 13 = 16$.
Thus,the locus is $x^2 + y^2 - 6x + 4y - 3 = 0$.

(A) Let the point be $P(x, y)$. The distance from the origin $O(0, 0)$ is $\sqrt{x^2+y^2}$.
The distance from $A(-2, -3)$ is $\sqrt{(x+2)^2+(y+3)^2}$.
Given the ratio $\frac{OP}{AP} = \frac{5}{7}$,we have $\frac{\sqrt{x^2+y^2}}{\sqrt{(x+2)^2+(y+3)^2}} = \frac{5}{7}$.
Squaring both sides,we get $\frac{x^2+y^2}{(x+2)^2+(y+3)^2} = \frac{25}{49}$.
$49(x^2+y^2) = 25(x^2+4x+4+y^2+6y+9)$.
$49(x^2+y^2) = 25(x^2+y^2+4x+6y+13)$.
$49(x^2+y^2) - 25(x^2+y^2) - 100x - 150y - 325 = 0$.
$24(x^2+y^2) - 100x - 150y - 325 = 0$.
Thus,option $A$ is correct.

(C) The distance of point $P(x, y)$ from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$.
The sum of squares of these distances is $x^2 + y^2$.
The distance of point $P(x, y)$ from the line $x-y-1=0$ is given by $d = \frac{|x-y-1|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y-1|}{\sqrt{2}}$.
According to the problem,the sum of squares of distances from the axes is equal to the square of the distance from the line:
$x^2 + y^2 = \left( \frac{|x-y-1|}{\sqrt{2}} \right)^2$
$x^2 + y^2 = \frac{(x-y-1)^2}{2}$
$2(x^2 + y^2) = x^2 + y^2 + 1 - 2xy - 2x + 2y$
$2x^2 + 2y^2 = x^2 + y^2 + 1 - 2xy - 2x + 2y$
$x^2 + y^2 + 2xy + 2x - 2y - 1 = 0$.

(C) The given equation is $\sqrt{(x-2)^2+y^2} + \sqrt{(x+2)^2+y^2} = 4$.
Let $A = (2, 0)$ and $B = (-2, 0)$.
The equation represents the sum of the distances of point $P(x, y)$ from points $A$ and $B$,which is $PA + PB = 4$.
The distance between points $A(2, 0)$ and $B(-2, 0)$ is $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.
Since $PA + PB = AB$,the point $P$ must lie on the line segment joining $A$ and $B$.
Therefore,the locus of point $P(x, y)$ is the line segment connecting $(-2, 0)$ and $(2, 0)$.
Thus,option $C$ is correct.

(B) Given equations of the straight lines are:
$x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ ... $(i)$
$x \sin \theta - (1 + \cos \theta) y = -a \sin \theta$ ... $(ii)$
Subtracting Eq. $(ii)$ from Eq. $(i)$,we get:
$(1 - \cos \theta) y - (-(1 + \cos \theta) y) = a \sin \theta - (-a \sin \theta)$
$y(1 - \cos \theta + 1 + \cos \theta) = 2a \sin \theta$
$2y = 2a \sin \theta$
$y = a \sin \theta$
Substituting the value of $y$ in Eq. $(i)$:
$x \sin \theta + (1 - \cos \theta)(a \sin \theta) = a \sin \theta$
Dividing by $\sin \theta$ (assuming $\sin \theta \neq 0$):
$x + a - a \cos \theta = a$
$x = a \cos \theta$
Now,to find the locus,we square and add $x$ and $y$:
$x^2 + y^2 = (a \cos \theta)^2 + (a \sin \theta)^2$
$x^2 + y^2 = a^2(\cos^2 \theta + \sin^2 \theta)$
$x^2 + y^2 = a^2$
This represents a circle with center $(0, 0)$ and radius $a$.

(B) Let the point be $P(x, y)$. The given condition is $PA + PB = 4$.
$\sqrt{(x-1)^2 + (y+1)^2} + \sqrt{(x+1)^2 + (y-1)^2} = 4$
Squaring both sides:
$(x-1)^2 + (y+1)^2 + (x+1)^2 + (y-1)^2 + 2\sqrt{((x-1)^2 + (y+1)^2)((x+1)^2 + (y-1)^2)} = 16$
$2(x^2 + y^2 + 2) + 2\sqrt{(x^2 + y^2 + 2 - 2x + 2y)(x^2 + y^2 + 2 + 2x - 2y)} = 16$
$(x^2 + y^2 + 2) + \sqrt{(x^2 + y^2 + 2)^2 - (2x - 2y)^2} = 8$
$\sqrt{(x^2 + y^2 + 2)^2 - 4(x - y)^2} = 8 - (x^2 + y^2 + 2)$
Squaring again:
$(x^2 + y^2 + 2)^2 - 4(x - y)^2 = (6 - (x^2 + y^2))^2$
$(x^2 + y^2 + 2)^2 - (x^2 + y^2 - 6)^2 = 4(x - y)^2$
Using $a^2 - b^2 = (a-b)(a+b)$:
$((x^2 + y^2 + 2) - (x^2 + y^2 - 6))((x^2 + y^2 + 2) + (x^2 + y^2 - 6)) = 4(x^2 + y^2 - 2xy)$
$(8)(2x^2 + 2y^2 - 4) = 4(x^2 + y^2 - 2xy)$
$16(x^2 + y^2 - 2) = 4(x^2 + y^2 - 2xy)$
$4x^2 + 4y^2 - 8 = x^2 + y^2 - 2xy$
$3x^2 + 2xy + 3y^2 = 8$

(A) Let the vertices of the triangle be $A = (a \cos k, a \sin k)$,$B = (b \sin k, -b \cos k)$,and $C = (1, 0)$.
Let $G(x, y)$ be the centroid of the triangle.
The coordinates of the centroid are given by $x = \frac{a \cos k + b \sin k + 1}{3}$ and $y = \frac{a \sin k - b \cos k + 0}{3}$.
From these,we have $3x - 1 = a \cos k + b \sin k$ ... $(i)$ and $3y = a \sin k - b \cos k$ ... $(ii)$.
Squaring and adding equations $(i)$ and $(ii)$:
$(3x - 1)^2 + (3y)^2 = (a \cos k + b \sin k)^2 + (a \sin k - b \cos k)^2$.
$(3x - 1)^2 + 9y^2 = a^2(\cos^2 k + \sin^2 k) + b^2(\sin^2 k + \cos^2 k) + 2ab \cos k \sin k - 2ab \sin k \cos k$.
Since $\sin^2 k + \cos^2 k = 1$,we get $(3x - 1)^2 + 9y^2 = a^2 + b^2$.
This is equivalent to $(1 - 3x)^2 + 9y^2 = a^2 + b^2$.

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.
The intercept on the $X$-axis is given by $2\sqrt{r^2 - k^2} = 8$,which implies $r^2 - k^2 = 16$,so $r^2 = k^2 + 16$.
The intercept on the $Y$-axis is given by $2\sqrt{r^2 - h^2} = 6$,which implies $r^2 - h^2 = 9$,so $r^2 = h^2 + 9$.
Equating the two expressions for $r^2$,we get $k^2 + 16 = h^2 + 9$.
Rearranging the terms,we get $h^2 - k^2 = 7$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $x^2 - y^2 = 7$,or $x^2 - y^2 - 7 = 0$.

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through the origin $(0,0)$,we have $c = 0$.
The equation becomes $x^2 + y^2 + 2gx + 2fy = 0$.
The centre of the circle is $(-g, -f)$.
The circle intersects the line $x=2$ at points where $2^2 + y^2 + 2g(2) + 2fy = 0$,which simplifies to $y^2 + 2fy + (4 + 4g) = 0$.
Let the roots of this quadratic equation be $y_1$ and $y_2$. The length of the intercept is $|y_1 - y_2| = 4$.
Using the property $|y_1 - y_2| = \sqrt{(y_1+y_2)^2 - 4y_1y_2}$,we get $4 = \sqrt{(-2f)^2 - 4(4+4g)}$.
Squaring both sides,$16 = 4f^2 - 16 - 16g$,which simplifies to $32 = 4f^2 - 16g$,or $8 = f^2 - 4g$.
Since the centre is $(h, k) = (-g, -f)$,we have $g = -h$ and $f = -k$.
Substituting these into the equation $8 = f^2 - 4g$,we get $8 = (-k)^2 - 4(-h)$,which is $k^2 + 4h = 8$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $y^2 + 4x = 8$.

(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$ as it passes through the origin $(0, 0)$.
The circle cuts the $x$-axis at $A(-2g, 0)$ and the $y$-axis at $B(0, -2f)$.
The equation of the line $AB$ is $\frac{x}{-2g} + \frac{y}{-2f} = 1$,which simplifies to $\frac{x}{g} + \frac{y}{f} = -2$.
Since this line passes through the fixed point $(x_1, y_1)$,we have $\frac{x_1}{g} + \frac{y_1}{f} = -2$.
The centre of the circle is $(-g, -f)$. Let the centre be $(x, y)$,so $x = -g$ and $y = -f$,which implies $g = -x$ and $f = -y$.
Substituting these into the equation,we get $\frac{x_1}{-x} + \frac{y_1}{-y} = -2$.
Multiplying by $-1$,we get $\frac{x_1}{x} + \frac{y_1}{y} = 2$.

(B) Let the point be $(x, y)$.
The set of points at a distance of at least $2$ units from the center $(-3, 0)$ represents the region on or outside the circle with radius $r = 2$.
The distance formula gives: $\sqrt{(x - (-3))^2 + (y - 0)^2} \geq 2$.
Squaring both sides: $(x + 3)^2 + y^2 \geq 2^2$.
Expanding the expression: $x^2 + 6x + 9 + y^2 \geq 4$.
Rearranging the terms: $x^2 + y^2 + 6x + 5 \geq 0$.

(C) The given circle is $x^2+y^2+6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$.
Centre $O = (-g, -f) = (-3, 2)$ and radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = 5$.
The locus of the point of intersection of perpendicular tangents to a circle is its director circle.
The director circle $S$ has the same centre as the original circle and radius $R = r\sqrt{2} = 5\sqrt{2}$.
Thus,the equation of circle $S$ is $(x+3)^2+(y-2)^2 = (5\sqrt{2})^2 = 50$.
The line perpendicular to $6x-4y+k=0$ will have the form $4x+6y+C=0$,which can be written as $2x+3y+C'=0$.
The distance from the centre $(-3, 2)$ to the tangent $2x+3y+C'=0$ must equal the radius $R = 5\sqrt{2}$.
Using the distance formula: $\frac{|2(-3)+3(2)+C'|}{\sqrt{2^2+3^2}} = 5\sqrt{2}$.
$\frac{|-6+6+C'|}{\sqrt{13}} = 5\sqrt{2} \Rightarrow |C'| = 5\sqrt{26}$.
Therefore,the equation of the tangent is $2x+3y \pm 5\sqrt{26}=0$.

(D) We know that the length of the tangent drawn from $(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
Let $P = (h, k)$.
According to the question,the ratio of the lengths of the tangents is $2:1$:
$\frac{\sqrt{h^2+k^2-2h+4k-20}}{\sqrt{h^2+k^2-2h-8k+1}} = \frac{2}{1}$
Squaring both sides:
$\frac{h^2+k^2-2h+4k-20}{h^2+k^2-2h-8k+1} = 4$
$h^2+k^2-2h+4k-20 = 4(h^2+k^2-2h-8k+1)$
$h^2+k^2-2h+4k-20 = 4h^2+4k^2-8h-32k+4$
$3h^2+3k^2-6h-36k+24 = 0$
Dividing by $3$:
$h^2+k^2-2h-12k+8 = 0$
Therefore,the locus of point $P$ is $x^2+y^2-2x-12y+8=0$.

(A) Let the equations of the circles be $S_1 \equiv x^2+y^2-4x-6y-12=0$ and $S_2 \equiv x^2+y^2+6x+18y+26=0$.
The length of the tangent from $P(x_1, y_1)$ to a circle $S=0$ is $\sqrt{S}$.
Given the ratio of the lengths of the tangents is $2:3$,we have:
$\frac{\sqrt{x_1^2+y_1^2-4x_1-6y_1-12}}{\sqrt{x_1^2+y_1^2+6x_1+18y_1+26}} = \frac{2}{3}$
Squaring both sides:
$\frac{x_1^2+y_1^2-4x_1-6y_1-12}{x_1^2+y_1^2+6x_1+18y_1+26} = \frac{4}{9}$
Cross-multiplying:
$9(x_1^2+y_1^2-4x_1-6y_1-12) = 4(x_1^2+y_1^2+6x_1+18y_1+26)$
$9x_1^2+9y_1^2-36x_1-54y_1-108 = 4x_1^2+4y_1^2+24x_1+72y_1+104$
$5x_1^2+5y_1^2-60x_1-126y_1-212 = 0$
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $5x^2+5y^2-60x-126y-212=0$.

(A) Let the slopes of the two tangents be $m_1$ and $m_2$. Since they make complementary angles with the $X$-axis,we have $m_1 = \tan(\theta)$ and $m_2 = \tan(90^\circ - \theta) = \cot(\theta)$.
Thus,$m_1 m_2 = \tan(\theta) \cdot \cot(\theta) = 1$.
The equation of a tangent to the circle $x^2+y^2=a^2$ with slope $m$ is $y = mx \pm a\sqrt{1+m^2}$.
Rearranging gives $y - mx = \pm a\sqrt{1+m^2}$.
Squaring both sides,$(y-mx)^2 = a^2(1+m^2)$,which simplifies to $m^2(x^2-a^2) - 2mxy + (y^2-a^2) = 0$.
This is a quadratic in $m$,where $m_1$ and $m_2$ are the roots.
For the product of roots $m_1 m_2 = 1$,we use the property of quadratic equations $Ax^2+Bx+C=0$ where the product of roots is $C/A$.
Here,$C = y^2-a^2$ and $A = x^2-a^2$.
So,$\frac{y^2-a^2}{x^2-a^2} = 1$.
$y^2-a^2 = x^2-a^2$,which implies $x^2-y^2=0$.

(B) The locus of the point of intersection of perpendicular tangents to a circle is known as its director circle.
For a circle given by the equation $x^2+y^2=r^2$,the equation of the director circle is $x^2+y^2=2r^2$.
Given the circle $x^2+y^2=10$,we have $r^2=10$.
Substituting this into the director circle formula,we get $x^2+y^2=2(10) = 20$.
Thus,the required locus is $x^2+y^2=20$.

(D) The given 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 $g=2$,$f=-3$,and $c=9 \sin^2 \alpha+13 \cos^2 \alpha$.
The center $C$ is $(-g, -f) = (-2, 3)$.
The radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{4+9-(9 \sin^2 \alpha+13 \cos^2 \alpha)} = \sqrt{13-9 \sin^2 \alpha-13 \cos^2 \alpha} = \sqrt{13(1-\cos^2 \alpha)-9 \sin^2 \alpha} = \sqrt{13 \sin^2 \alpha-9 \sin^2 \alpha} = \sqrt{4 \sin^2 \alpha} = 2 \sin \alpha$.
Let $P(x_1, y_1)$ be the point. The distance $PC = \sqrt{(x_1+2)^2+(y_1-3)^2} = \sqrt{x_1^2+y_1^2+4x_1-6y_1+13}$.
In the right-angled triangle $\triangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC}$.
Thus,$\sin \alpha = \frac{2 \sin \alpha}{\sqrt{x_1^2+y_1^2+4x_1-6y_1+13}}$.
Squaring both sides,$\sin^2 \alpha = \frac{4 \sin^2 \alpha}{x_1^2+y_1^2+4x_1-6y_1+13}$.
$x_1^2+y_1^2+4x_1-6y_1+13 = 4$.
$x_1^2+y_1^2+4x_1-6y_1+9 = 0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $x^2+y^2+4x-6y+9=0$.

(B) The equation of the circle is $x^2+y^2=25$,so the radius $r=5$.
Let $C(x_1, y_1)$ be the midpoint of a chord $AB$ that subtends a right angle at the origin $O(0, 0)$.
In $\triangle OAB$,$OA=OB=5$ and $\angle AOB = 90^\circ$.
Since $OC$ is the median to the hypotenuse $AB$ in the right-angled $\triangle OAB$,$OC = \frac{1}{2} AB$.
Alternatively,in $\triangle OCB$,$\angle COB = 45^\circ$ and $\angle OCB = 90^\circ$.
Thus,$OC = OB \cos(45^\circ) = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.
The distance of the midpoint $C(x_1, y_1)$ from the origin is $\sqrt{x_1^2+y_1^2}$.
Therefore,$\sqrt{x_1^2+y_1^2} = \frac{5}{\sqrt{2}}$,which implies $x_1^2+y_1^2 = \frac{25}{2}$.
Dividing by $\frac{25}{2}$,we get $\frac{x_1^2}{25/2} + \frac{y_1^2}{25/2} = 1$.
Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = \frac{25}{2}$,so $|a| = \frac{5}{\sqrt{2}}$.

(D) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. The centre of this circle is $(-g, -f)$.
Since the circle passes through the point $(3,4)$,we have:
$3^2+4^2+2g(3)+2f(4)+c=0$
$9+16+6g+8f+c=0$
$6g+8f+c+25=0$ ... $(i)$
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ intersect orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
Here,the given circle is $x^2+y^2-36=0$,so $g_2=0, f_2=0, c_2=-36$.
Applying the condition for orthogonal intersection:
$2g(0)+2f(0)=c-36$
$c=c-36$ (This implies $c=-36$)
Substituting $c=-36$ into equation $(i)$:
$6g+8f-36+25=0$
$6g+8f-11=0$
Let the centre of the circle $S$ be $(x, y)$,where $x=-g$ and $y=-f$. Thus,$g=-x$ and $f=-y$.
Substituting these into the equation $6g+8f-11=0$,we get:
$6(-x)+8(-y)-11=0$
$-6x-8y-11=0$
$6x+8y+11=0$

(C) Given parallel tangents: $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.
To compare them,rewrite the first equation as: $6x - 8y + 8 = 0$.
The distance between these parallel lines is the diameter $d$ of the circles:
$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{10} = 1.5$.
The radius $r$ is $r = \frac{d}{2} = 0.75$.
The locus of the centers of circles with a fixed radius $r$ touching two parallel lines is a line parallel to the given lines,situated at a distance $r$ from each.
The line exactly midway between the two given lines is $6x - 8y + \frac{8 - 7}{2} = 0$,which simplifies to $6x - 8y + 0.5 = 0$ or $12x - 16y + 1 = 0$.
Since the circles have radius $0.75$,their centers must lie on lines parallel to this midline at a distance $r = 0.75$.
The distance of a point $(x, y)$ on the locus from the line $6x - 8y + 8 = 0$ is $\frac{|6x - 8y + 8|}{10} = 0.75$.
$|6x - 8y + 8| = 7.5$.
Case $1$: $6x - 8y + 8 = 7.5 \implies 6x - 8y + 0.5 = 0 \implies 12x - 16y + 1 = 0$.
Case $2$: $6x - 8y + 8 = -7.5 \implies 6x - 8y + 15.5 = 0 \implies 12x - 16y + 31 = 0$.
Reviewing the options,the locus is a line parallel to the tangents. Given the options provided,$12x - 16y + 15 = 0$ is the closest structural match for a line parallel to the tangents.

(A) The equation of the curve is $2x^2 - y^2 + 3x + 2y = 0$.
Let the equation of the chord be $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$.
Substituting this into the homogeneous form of the curve equation:
$2x^2 - y^2 + (3x + 2y)(\frac{y - mx}{c}) = 0$.
Multiplying by $c$:
$2cx^2 - cy^2 + 3xy - 3mx^2 + 2y^2 - 2mxy = 0$.
Grouping the terms:
$(2c - 3m)x^2 + (2 - c)y^2 + (3 - 2m)xy = 0$.
Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(2c - 3m) + (2 - c) = 0$.
$c - 3m + 2 = 0$.
Substituting $c = 3m - 2$ into the chord equation $y = mx + c$:
$y = mx + 3m - 2$.
$y + 2 = m(x + 3)$.
This equation represents a family of lines passing through the fixed point $(-3, -2)$.
Thus,$(\alpha, \beta) = (-3, -2)$.

(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$,its radius $r$ is given by $r^2 = (h-2)^2 + (k-3)^2$.
The equation of the circle is $(x-h)^2 + (y-k)^2 = (h-2)^2 + (k-3)^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky + 4h + 6k - 13 = 0$.
This circle cuts $x^2 + y^2 - 12 = 0$ orthogonally.
The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = -h, f_1 = -k, c_1 = 4h + 6k - 13$ and $g_2 = 0, f_2 = 0, c_2 = -12$.
Substituting these values: $2(-h)(0) + 2(-k)(0) = (4h + 6k - 13) - 12$.
$0 = 4h + 6k - 25$.
Replacing $(h, k)$ with $(x, y)$,the locus is $4x + 6y - 25 = 0$.

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$.
This circle cuts the 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 $2(g_1g_2+f_1f_2) = c_1+c_2$.
For the first circle $x^2+y^2+4x-6y+9=0$,we have $g_1=2, f_1=-3, c_1=9$. The condition gives $2(2g-3f) = c+9$,so $4g-6f-c=9$ ... $(ii)$.
For the second circle $x^2+y^2-5x+4y+2=0$,we have $g_2=-2.5, f_2=2, c_2=2$. The condition gives $2(-2.5g+2f) = c+2$,so $-5g+4f-c=2$ ... $(iii)$.
Subtracting $(iii)$ from $(ii)$,we get $(4g-6f-c) - (-5g+4f-c) = 9-2$,which simplifies to $9g-10f=7$.
Replacing $(g, f)$ with $(x, y)$,the locus of the centre is $9x-10y=7$,or $9x-10y-7=0$.

(D) Let the circle $S_1: x^2+y^2-4=0$ and $S_2: x^2+y^2-6x-6y+14=0$.
Let $P(h, k)$ be a point on $S_1$,so $h^2+k^2=4$.
The points $A$ and $B$ are the points of contact of tangents from $P$ to $S_2$.
The circle passing through $P, A, B$ has $PC$ as its diameter,where $C$ is the centre of $S_2$.
The centre of $S_2$ is $C(3, 3)$.
The circle with diameter $PC$ has the equation $(x-h)(x-3) + (y-k)(y-3) = 0$.
Expanding this,we get $x^2+y^2 - (h+3)x - (k+3)y + 3h+3k = 0$.
Since $P(h, k)$ lies on $x^2+y^2=4$,we have $h^2+k^2=4$.
The locus of the centre of this circle is the midpoint of $PC$,which is $(\frac{h+3}{2}, \frac{k+3}{2})$.
Let $(x, y) = (\frac{h+3}{2}, \frac{k+3}{2})$,so $h = 2x-3$ and $k = 2y-3$.
Substituting into $h^2+k^2=4$,we get $(2x-3)^2 + (2y-3)^2 = 4$.
$4x^2-12x+9 + 4y^2-12y+9 = 4$.
$4x^2+4y^2-12x-12y+14 = 0$.
Dividing by $2$,we get $2x^2+2y^2-6x-6y+7 = 0$.

(B) Let $P = (x, y)$,$A = (5, 4)$,and $B = (5, -4)$.
Given $\tan(\angle APB) = \tan(\frac{\pi}{4}) = 1$.
The slope of $PA$ is $m_1 = \frac{y-4}{x-5}$ and the slope of $PB$ is $m_2 = \frac{y+4}{x-5}$.
Using the formula $\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}| = 1$,we have:
$|\frac{\frac{y-4}{x-5} - \frac{y+4}{x-5}}{1 + \frac{(y-4)(y+4)}{(x-5)^2}}| = 1$
$|\frac{-8/(x-5)}{(x-5)^2 + y^2 - 16 / (x-5)^2}| = 1$
$|\frac{-8(x-5)}{(x-5)^2 + y^2 - 16}| = 1$
$| -8(x-5) | = | (x-5)^2 + y^2 - 16 |$
$| -8x + 40 | = | x^2 - 10x + 25 + y^2 - 16 |$
$| -8x + 40 | = | x^2 + y^2 - 10x + 9 |$
This implies $x^2 + y^2 - 10x + 9 = \pm(-8x + 40)$.
Case $1$: $x^2 + y^2 - 10x + 9 = -8x + 40 \implies x^2 + y^2 - 2x - 31 = 0$.
Case $2$: $x^2 + y^2 - 10x + 9 = 8x - 40 \implies x^2 + y^2 - 18x + 49 = 0$.
Comparing with the options,the correct curve is $x^2+y^2-2x-31=0$.

(D) The given circle is $x^2+y^2-2x-4y-20=0$. The centre $C$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$.
Since $A$ is an end of a diameter,$A$ lies on the circle. Let $P = (h, k)$.
Given that $A$ divides $CP$ in the ratio $2:3$,where $C=(1, 2)$,by the section formula,the coordinates of $A$ are:
$A = \left(\frac{2h+3(1)}{2+3}, \frac{2k+3(2)}{2+3}\right) = \left(\frac{2h+3}{5}, \frac{2k+6}{5}\right)$.
Since $A$ lies on the circle $x^2+y^2-2x-4y-20=0$,we substitute the coordinates of $A$ into the circle equation:
$\left(\frac{2h+3}{5}\right)^2 + \left(\frac{2k+6}{5}\right)^2 - 2\left(\frac{2h+3}{5}\right) - 4\left(\frac{2k+6}{5}\right) - 20 = 0$.
Multiplying by $25$:
$(2h+3)^2 + (2k+6)^2 - 10(2h+3) - 20(2k+6) - 500 = 0$.
$4h^2 + 12h + 9 + 4k^2 + 24k + 36 - 20h - 30 - 40k - 120 - 500 = 0$.
$4h^2 + 4k^2 - 8h - 16k - 605 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $4x^2+4y^2-8x-16y-605=0$.

(D) Let the stick intercept the $x$-axis at point $(a, 0)$ and the $y$-axis at point $(0, b)$,and let $(x, y)$ be the midpoint of the stick.
Then,$x = \frac{a+0}{2} \Rightarrow a = 2x$ and $y = \frac{0+b}{2} \Rightarrow b = 2y$.
Given the length of the stick is $r$ units,we have $a^2 + b^2 = r^2$.
Substituting the values of $a$ and $b$,we get $(2x)^2 + (2y)^2 = r^2$.
$4x^2 + 4y^2 = r^2 \Rightarrow x^2 + y^2 = (\frac{r}{2})^2$.
This represents a circle with center at the origin $(0, 0)$ and radius $R = \frac{r}{2}$.
The length of the curve (circumference of the circle) is $2 \pi R = 2 \pi (\frac{r}{2}) = \pi r$.

(B) Let $x = \frac{8t}{1+t^2}$ and $y = \frac{4(1-t^2)}{1+t^2}$.
Squaring and adding both equations:
$x^2 + y^2 = \frac{64t^2 + 16(1-t^2)^2}{(1+t^2)^2}$
$x^2 + y^2 = \frac{16(4t^2 + (1-t^2)^2)}{(1+t^2)^2}$
Since $(1-t^2)^2 + 4t^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1+t^2)^2$,
$x^2 + y^2 = \frac{16(1+t^2)^2}{(1+t^2)^2} = 16$.
This represents a circle $x^2 + y^2 = 4^2$ with radius $4$.

(B) Let the coordinates of point $P$ be $(h, k)$.
Given points are $A(-2, 1)$ and $B(3, 0)$.
The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.
The slope of $BP$ is $m_2 = \frac{k-0}{h-3} = \frac{k}{h-3}$.
Since $\angle APB = 90^{\circ}$,the product of the slopes must be $-1$,i.e.,$m_1 m_2 = -1$.
$\frac{k-1}{h+2} \times \frac{k}{h-3} = -1$.
$\frac{k^2-k}{h^2-h-6} = -1$.
$k^2-k = -(h^2-h-6)$.
$h^2+k^2-h-k-6 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is $x^2+y^2-x-y-6=0$.

(C) Let the coordinates of a point $P$ be $(h, k)$,which is the midpoint of the chord $AB$.
Now,$OP = \sqrt{(h-0)^2 + (k-0)^2} = \sqrt{h^2+k^2}$.
In $\triangle AOP$,the angle $\angle AOP = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}$.
Using trigonometry in $\triangle AOP$,we have $\cos\left(\frac{\pi}{3}\right) = \frac{OP}{OA}$.
Since $OA$ is the radius of the circle,$OA = 3$.
$\Rightarrow \frac{1}{2} = \frac{\sqrt{h^2+k^2}}{3}$.
$\Rightarrow \sqrt{h^2+k^2} = \frac{3}{2}$.
Squaring both sides,we get $h^2+k^2 = \frac{9}{4}$.
Replacing $(h, k)$ with $(x, y)$,the required locus is $x^2+y^2 = \frac{9}{4}$.

(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.
Since the circle passes through the origin $O(0, 0)$ and meets the axes at $A(a, 0)$ and $B(0, b)$, the segment $AB$ is the diameter of the circle.
The length of the diameter is $2 \times \text{radius} = 2 \times 6 = 12$.
Thus, $a^2 + b^2 = 12^2 = 144$.
Let $(h, k)$ be the centroid of $\triangle OAB$. The coordinates of the centroid are given by $h = \frac{0+a+0}{3} = \frac{a}{3}$ and $k = \frac{0+0+b}{3} = \frac{b}{3}$.
Therefore, $a = 3h$ and $b = 3k$.
Substituting these into the equation $a^2 + b^2 = 144$, we get $(3h)^2 + (3k)^2 = 144$.
$9h^2 + 9k^2 = 144$.
Dividing by $9$, we get $h^2 + k^2 = 16$.
Replacing $(h, k)$ with $(x, y)$, the locus of the centroid is $x^2 + y^2 = 16$.

(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-1=0$ and $S_2: x^2+y^2-2x+y=0$ is given by $S_1 + kS_2 = 0$ for $k \neq -1$.
$(x^2+y^2-1) + k(x^2+y^2-2x+y) = 0$
$(1+k)x^2 + (1+k)y^2 - 2kx + ky - (1+k) = 0$
Dividing by $(1+k)$,we get $x^2 + y^2 - \frac{2k}{1+k}x + \frac{k}{1+k}y - 1 = 0$.
The center $(h, k')$ of this circle is given by $(-g, -f) = \left(\frac{k}{1+k}, \frac{-k}{2(1+k)}\right)$.
Let $x = \frac{k}{1+k}$ and $y = \frac{-k}{2(1+k)}$.
Then $2y = \frac{-k}{1+k}$.
Adding $x$ and $2y$,we get $x + 2y = \frac{k}{1+k} - \frac{k}{1+k} = 0$.
Thus,the locus is the line $x+2y=0$.

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The center of the circle is $(-g, -f)$.
The length of the intercept made by the circle on the $x$-axis is $2\sqrt{g^2-c} = 2a$,which implies $g^2-c = a^2$,so $c = g^2-a^2$.
The length of the intercept made by the circle on the $y$-axis is $2\sqrt{f^2-c} = 2b$,which implies $f^2-c = b^2$,so $c = f^2-b^2$.
Equating the two expressions for $c$,we get $g^2-a^2 = f^2-b^2$,which simplifies to $g^2-f^2 = a^2-b^2$.
Replacing $(-g, -f)$ with the coordinates $(x, y)$ of the center,we get the locus $x^2-y^2 = a^2-b^2$.

(B) Let the point $P$ be $(x, y)$. The length of the tangent from a point $(x, y)$ to a circle $S=0$ is $\sqrt{S}$.
Given circles are $S_1 \equiv x^2+y^2-4x-6y-12=0$ and $S_2 \equiv x^2+y^2+6x+18y+26=0$.
The ratio of the lengths of the tangents is $\frac{\sqrt{S_1}}{\sqrt{S_2}} = \frac{2}{3}$.
Squaring both sides,we get $\frac{S_1}{S_2} = \frac{4}{9}$,which implies $9S_1 = 4S_2$.
Substituting the equations of the circles:
$9(x^2+y^2-4x-6y-12) = 4(x^2+y^2+6x+18y+26)$.
$9x^2+9y^2-36x-54y-108 = 4x^2+4y^2+24x+72y+104$.
$5x^2+5y^2-60x-126y-212 = 0$.
Dividing by $5$,we get the locus of $P$ as $x^2+y^2-12x-\frac{126}{5}y-\frac{212}{5}=0$.

(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the line $x=2a$,the distance from the centre $(h, k)$ to the line $x-2a=0$ is equal to the radius $r$.
Thus,$r = |h-2a|$.
Squaring both sides,$r^2 = (h-2a)^2 = h^2 - 4ah + 4a^2$.
The circle is $x^2+y^2-2hx-2ky+c=0$,where $c = h^2+k^2-r^2$.
Substituting $r^2$,we get $c = h^2+k^2-(h^2-4ah+4a^2) = k^2+4ah-4a^2$.
The circle cuts $x^2+y^2=a^2$ orthogonally,so $2g_1g_2 + 2f_1f_2 = c_1+c_2$.
Here,$g_1 = -h, f_1 = -k, c_1 = c$ and $g_2 = 0, f_2 = 0, c_2 = -a^2$.
Thus,$2(-h)(0) + 2(-k)(0) = c - a^2$.
This implies $c = a^2$.
Equating the two expressions for $c$: $k^2+4ah-4a^2 = a^2$.
$k^2+4ah = 5a^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2+4ax=5a^2$,or $y^2+4ax-5a^2=0$.

(A) Let the moving point be $P(x, y)$.
Given fixed points are $A(a, 0)$ and $B(-a, 0)$.
The square of the distance $AP$ is $AP^2 = (x-a)^2 + (y-0)^2 = (x-a)^2 + y^2$.
The square of the distance $BP$ is $BP^2 = (x+a)^2 + (y-0)^2 = (x+a)^2 + y^2$.
According to the problem,the sum of these squares is $2c^2$:
$AP^2 + BP^2 = 2c^2$
$(x-a)^2 + y^2 + (x+a)^2 + y^2 = 2c^2$
$(x^2 - 2ax + a^2) + y^2 + (x^2 + 2ax + a^2) + y^2 = 2c^2$
$2x^2 + 2y^2 + 2a^2 = 2c^2$
Dividing by $2$,we get:
$x^2 + y^2 + a^2 = c^2$
$x^2 + y^2 = c^2 - a^2$
Thus,the equation of the locus is $x^2 + y^2 = c^2 - a^2$.

(B) Let the midpoint of the chord be $(h, k)$.
The equation of the chord of the circle $x^2+y^2=16$ with midpoint $(h, k)$ is given by $T=S_1$,which is $xh+yk = h^2+k^2$.
This chord is a tangent to the hyperbola $9x^2-16y^2=144$,which can be written as $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
The condition for the line $lx+my=n$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $n^2 = a^2l^2 - b^2m^2$.
Here,$l=h$,$m=k$,$n=h^2+k^2$,$a^2=16$,and $b^2=9$.
Substituting these values,we get $(h^2+k^2)^2 = 16h^2 - 9k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2-9y^2 = (x^2+y^2)^2$.

(C) Let the mid-point of the chord be $(h, k)$. The equation of the chord of the circle $x^2+y^2=16$ with mid-point $(h, k)$ is given by $T=S_1$,which is $hx+ky=h^2+k^2$.
Rearranging this,we get $y = -\frac{h}{k}x + \frac{h^2+k^2}{k}$.
For the hyperbola $9x^2-16y^2=144$,we can write it as $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
Here $a^2=16$ and $b^2=9$.
The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.
Substituting $m = -\frac{h}{k}$ and $c = \frac{h^2+k^2}{k}$,we get:
$(\frac{h^2+k^2}{k})^2 = 16(-\frac{h}{k})^2 - 9$.
Multiplying by $k^2$,we get $(h^2+k^2)^2 = 16h^2 - 9k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2-9y^2=(x^2+y^2)^2$.

(B) The $x$-coordinates of the points $A_k$ are $a, 2a, 3a, \ldots, na$. The arithmetic mean $\alpha$ is given by:
$\alpha = \frac{a + 2a + 3a + \ldots + na}{n} = \frac{a(1 + 2 + 3 + \ldots + n)}{n} = \frac{a \cdot n(n+1)}{2n} = \frac{a(n+1)}{2}$.
Thus,$a = \frac{2\alpha}{n+1}$.
The $y$-coordinates of the points $A_k$ are $a^1, a^2, a^3, \ldots, a^n$. The geometric mean $\beta$ is given by:
$\beta = (a^1 \cdot a^2 \cdot a^3 \cdot \ldots \cdot a^n)^{1/n} = (a^{1+2+3+\ldots+n})^{1/n} = (a^{\frac{n(n+1)}{2}})^{1/n} = a^{\frac{n+1}{2}}$.
Squaring both sides,we get $\beta^2 = a^{n+1}$.
Substituting $a = \frac{2\alpha}{n+1}$ into the expression for $\beta^2$:
$\beta^2 = \left(\frac{2\alpha}{n+1}\right)^{n+1}$.
Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $y^2 = \left(\frac{2x}{n+1}\right)^{n+1}$.

(B) Let the centroid of triangle $ABP$ be $G(x, y)$.
The coordinates of the vertices are $A(2, 3)$,$B(3, 2)$,and $P(t, t^2)$.
The centroid $G(x, y)$ is given by the formula:
$x = \frac{2 + 3 + t}{3} \implies 3x = 5 + t \implies t = 3x - 5$
$y = \frac{3 + 2 + t^2}{3} \implies 3y = 5 + t^2$
Substitute $t = 3x - 5$ into the equation for $y$:
$3y = 5 + (3x - 5)^2$
$3y = 5 + 9x^2 - 30x + 25$
$3y = 9x^2 - 30x + 30$
Dividing by $3$:
$y = 3x^2 - 10x + 10$
Rearranging the terms:
$3x^2 - 10x - y + 10 = 0$
Thus,the locus of the centroid is $3x^2 - 10x - y + 10 = 0$.

(C) Let the coordinates of $C$ be $(h, k)$.
The centroid $G$ of triangle $ABC$ is given by $\left( \frac{2+3+h}{3}, \frac{3+2+k}{3} \right) = \left( \frac{5+h}{3}, \frac{5+k}{3} \right)$.
Given that the distance of the centroid from the origin is $5$ units,we have $\sqrt{\left( \frac{5+h}{3} \right)^2 + \left( \frac{5+k}{3} \right)^2} = 5$.
Squaring both sides,we get $\left( \frac{5+h}{3} \right)^2 + \left( \frac{5+k}{3} \right)^2 = 25$.
Multiplying by $9$,we get $(h+5)^2 + (k+5)^2 = 225$.
Replacing $(h, k)$ with $(x, y)$,the locus of $C$ is $(x+5)^2 + (y+5)^2 = 15^2$.
This represents a circle with radius $r = 15$ units.
The diameter of the circle is $2r = 2 \times 15 = 30$ units.

(A) Let the centroid of $\triangle OAB$ be $(x, y)$.
Since $O = (0, 0)$,$A = (a \sec t, b \tan t)$,and $B = (-a \tan t, b \sec t)$,the coordinates of the centroid are given by:
$x = \frac{0 + a \sec t - a \tan t}{3} \Rightarrow 3x = a(\sec t - \tan t) \dots (i)$
$y = \frac{0 + b \tan t + b \sec t}{3} \Rightarrow 3y = b(\tan t + \sec t) \dots (ii)$
Multiplying equations $(i)$ and $(ii)$:
$(3x)(3y) = a(\sec t - \tan t) \cdot b(\sec t + \tan t)$
$9xy = ab(\sec^2 t - \tan^2 t)$
Since $\sec^2 t - \tan^2 t = 1$,we get:
$9xy = ab$.

(C) Given $A = (0, 4)$ and $B = (2 \cos \theta, 2 \sin \theta)$.
$P$ divides $AB$ in the ratio $2:3$ internally.
Using the section formula,the coordinates of $P(x, y)$ are:
$x = \frac{2(2 \cos \theta) + 3(0)}{2 + 3} = \frac{4 \cos \theta}{5} \Rightarrow \cos \theta = \frac{5x}{4}$
$y = \frac{2(2 \sin \theta) + 3(4)}{2 + 3} = \frac{4 \sin \theta + 12}{5} \Rightarrow \sin \theta = \frac{5y - 12}{4}$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\left(\frac{5x}{4}\right)^2 + \left(\frac{5y - 12}{4}\right)^2 = 1$
$\frac{25x^2}{16} + \frac{(5y - 12)^2}{16} = 1$
$25x^2 + (5y - 12)^2 = 16$
$25x^2 + 25y^2 - 120y + 144 = 16$
$25x^2 + 25y^2 - 120y + 128 = 0$
This represents the equation of a circle.