Geometrical problems regarding circle and its properties Questions in English

(D) Given,the slope of the diameter $AC$ is $m = \frac{3}{4}$.
Since $\tan \theta = \frac{3}{4}$,we have $\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$.
The centre of the circle is $(h, k) = (3, 2)$ and the radius is $r = 25$.
The coordinates of the endpoints of the diameter are given by $(h \pm r \cos \theta, k \pm r \sin \theta)$.
For $C = (x_2, y_2)$,using the positive sign:
$x_2 = 3 + 25 \times \frac{4}{5} = 3 + 20 = 23$
$y_2 = 2 + 25 \times \frac{3}{5} = 2 + 15 = 17$
For $A = (x_1, y_1)$,using the negative sign:
$x_1 = 3 - 25 \times \frac{4}{5} = 3 - 20 = -17$
$y_1 = 2 - 25 \times \frac{3}{5} = 2 - 15 = -13$
Therefore,$\frac{x_1 x_2}{y_1 y_2} = \frac{(-17) \times 23}{(-13) \times 17} = \frac{-17 \times 23}{-13 \times 17} = \frac{23}{13}$.

(B) Let the centre of the circle be $O = (x, y)$. Since the circle passes through $A(-1, 1)$,$B(2, -1)$,and $C(1, 0)$,we have $OA^2 = OB^2 = OC^2 = r^2$.
From $OA^2 = OB^2$:
$(x + 1)^2 + (y - 1)^2 = (x - 2)^2 + (y + 1)^2$
$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 4x + 4 + y^2 + 2y + 1$
$6x - 4y = 3$ ... $(i)$
From $OA^2 = OC^2$:
$(x + 1)^2 + (y - 1)^2 = (x - 1)^2 + y^2$
$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2$
$4x - 2y = -1$ ... $(ii)$
Multiplying $(ii)$ by $2$,we get $8x - 4y = -2$ ... $(iii)$
Subtracting $(i)$ from $(iii)$:
$2x = -5 \implies x = -\frac{5}{2}$
Substituting $x$ in $(ii)$:
$4(-\frac{5}{2}) - 2y = -1 \implies -10 - 2y = -1 \implies 2y = -9 \implies y = -\frac{9}{2}$
Centre $O = (-\frac{5}{2}, -\frac{9}{2})$.
Radius $r = \sqrt{(x - 1)^2 + (y - 0)^2} = \sqrt{(-\frac{5}{2} - 1)^2 + (-\frac{9}{2})^2} = \sqrt{(-\frac{7}{2})^2 + (-\frac{9}{2})^2} = \sqrt{\frac{49}{4} + \frac{81}{4}} = \sqrt{\frac{130}{4}} = \frac{\sqrt{130}}{2}$.

(A) The line divides the circumference in the ratio $1:2$,which corresponds to an arc angle of $\frac{1}{1+2} \times 360^{\circ} = 120^{\circ}$.
Let $O$ be the center $(5, 3)$ and $d$ be the perpendicular distance from $O$ to the line $4x + 3y - 4 = 0$.
$d = \frac{|4(5) + 3(3) - 4|}{\sqrt{4^2 + 3^2}} = \frac{|20 + 9 - 4|}{5} = \frac{25}{5} = 5$.
In the triangle formed by the center,the midpoint of the chord,and a point on the circumference,the angle at the center is half of the arc angle,i.e.,$60^{\circ}$.
Using trigonometry,$\cos(60^{\circ}) = \frac{d}{R}$,where $R$ is the radius.
$\frac{1}{2} = \frac{5}{R} \Rightarrow R = 10$.
The equation of the circle with center $(5, 3)$ and radius $10$ is $(x - 5)^2 + (y - 3)^2 = 10^2$.

(C) The given equations of the circles are:
$C_1: x^2+y^2+4x-6y-3=0$
$C_2: x^2+y^2+4x-2y+1=0$
Comparing with the general equation $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g=2, f=-3, c=-3$. Centre $O_1 = (-2, 3)$,Radius $r_1 = \sqrt{g^2+f^2-c} = \sqrt{4+9+3} = \sqrt{16} = 4$.
For $C_2$: $g=2, f=-1, c=1$. Centre $O_2 = (-2, 1)$,Radius $r_2 = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = \sqrt{4} = 2$.
The distance between the centres $O_1O_2 = \sqrt{(-2 - (-2))^2 + (3-1)^2} = \sqrt{0^2 + 2^2} = 2$.
We observe that $|r_1 - r_2| = |4 - 2| = 2$.
Since the distance between the centres $O_1O_2 = |r_1 - r_2|$,the two circles touch each other internally.
When two circles touch internally,there is exactly $1$ common tangent.

(D) The given circle is $x^2+y^2-2x-6y+6=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-3, c=6$.
The center of this circle is $(-g, -f) = (1, 3)$ and its radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+9-6} = \sqrt{4} = 2$.
$A$ diameter of this circle is a chord of the larger circle with center $O(2, 1)$.
The length of this chord is equal to the diameter of the smaller circle,which is $2r = 2(2) = 4$.
Let $A(1, 3)$ be the center of the smaller circle and $B$ be a point on the circumference of the smaller circle such that $AB$ is the radius $r=2$.
The distance $OA$ between the centers $(2, 1)$ and $(1, 3)$ is $OA = \sqrt{(2-1)^2 + (1-3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.
In the right-angled triangle formed by the center of the larger circle $O$,the center of the smaller circle $A$,and a point $B$ on the chord,we have $R^2 = OA^2 + r^2$,where $R$ is the radius of the larger circle.
$R^2 = (\sqrt{5})^2 + 2^2 = 5 + 4 = 9$.
Therefore,$R = \sqrt{9} = 3$.

(D) The power of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Given the circle equation $x^2 + y^2 + 4x - 6y - a = 0$ and the point $(1, 6)$,the power is $-16$.
Substituting the point $(1, 6)$ into the circle equation:
$(1)^2 + (6)^2 + 4(1) - 6(6) - a = -16$
$1 + 36 + 4 - 36 - a = -16$
$5 - a = -16$
$a = 5 + 16$
$a = 21$

(A) The given equation of the circle is $4x^2+4y^2-12x-12y+9=0$.
Dividing by $4$,we get $x^2+y^2-3x-3y+\frac{9}{4}=0$.
Rearranging the terms,we have $(x^2-3x)+(y^2-3y)=-\frac{9}{4}$.
Completing the square,we get $(x-\frac{3}{2})^2-\frac{9}{4}+(y-\frac{3}{2})^2-\frac{9}{4}=-\frac{9}{4}$.
This simplifies to $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=\frac{9}{4}$,which is $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=(\frac{3}{2})^2$.
Comparing this with the standard form $(x-h)^2+(y-k)^2=r^2$,the centre is $(\frac{3}{2}, \frac{3}{2})$ and the radius is $r=\frac{3}{2}$.
Since the distance of the centre from both axes is equal to the radius (i.e.,$|h|=|k|=r=\frac{3}{2}$),the circle touches both the axes.

(D) The equation of the circle is $x^2+y^2+2gx+2fy-23=0$. The centre is $(-g, -f) = (3, -2)$,so $g = -3$ and $f = 2$. The equation becomes $x^2+y^2-6x+4y-23=0$.
To find the radius $r$,$r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+2^2-(-23)} = \sqrt{9+4+23} = \sqrt{36} = 6$.
The point $A$ on the circle closest to $P(-1, -5)$ lies on the line segment connecting the centre $C(3, -2)$ and $P(-1, -5)$.
The vector $\vec{CP} = (-1-3, -5-(-2)) = (-4, -3)$.
The distance $CP = \sqrt{(-4)^2+(-3)^2} = 5$.
Since $CP < r$ $(5 < 6)$,the point $P$ lies inside the circle. The point $A$ on the circle closest to $P$ is the intersection of the line $CP$ with the circle,specifically the point such that $\vec{CA} = \frac{r}{CP} \vec{CP}$ in the direction of $\vec{CP}$.
However,for a point $P$ inside,the closest point $A$ is on the line $CP$ extended such that $A$ is on the circumference.
$A = C + \frac{r}{CP} \vec{CP} = (3, -2) + \frac{6}{5}(-4, -3) = (3 - \frac{24}{5}, -2 - \frac{18}{5}) = (-\frac{9}{5}, -\frac{28}{5})$.

(C) The given equation of the circle is $x^2+y^2-4x+4y+4=0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=4$.
The center $O$ is $(-g, -f) = (2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-4} = 2$.
The inverse point $Q(a, b)$ of $P(3, 3)$ lies on the line $OP$. The slope of $OP$ is $m = \frac{3 - (-2)}{3 - 2} = \frac{5}{1} = 5$.
The equation of line $OP$ is $y - (-2) = 5(x - 2)$,which simplifies to $y = 5x - 12$.
Since $Q(a, b)$ lies on this line,$b = 5a - 12$ (Equation $1$).
By the property of inverse points,$O, Q, P$ are collinear and $OQ \cdot OP = r^2$.
$OP = \sqrt{(3-2)^2 + (3-(-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$.
$OQ = \sqrt{(a-2)^2 + (b+2)^2} = \sqrt{(a-2)^2 + (5a-12+2)^2} = \sqrt{(a-2)^2 + (5a-10)^2} = \sqrt{(a-2)^2 + 25(a-2)^2} = \sqrt{26(a-2)^2} = |a-2|\sqrt{26}$.
Given $OQ \cdot OP = r^2$,we have $|a-2|\sqrt{26} \cdot \sqrt{26} = 2^2 = 4$.
$26|a-2| = 4 \Rightarrow |a-2| = \frac{4}{26} = \frac{2}{13}$.
Since $Q$ lies on the same side of $O$ as $P$,$a-2 = \frac{2}{13} \Rightarrow a = 2 + \frac{2}{13} = \frac{28}{13}$.
Then $b = 5(\frac{28}{13}) - 12 = \frac{140 - 156}{13} = -\frac{16}{13}$.
Finally,$a+5b = \frac{28}{13} + 5(-\frac{16}{13}) = \frac{28 - 80}{13} = -\frac{52}{13} = -4$.

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+C=0$.
Since the circle passes through $(-1,-1)$,we have $(-1)^2+(-1)^2+2g(-1)+2f(-1)+C=0$,which simplifies to $2-2g-2f+C=0$,or $2g+2f-C=2$ $... (i)$.
The power of the point $(2,0)$ with respect to the circle is given by $x^2+y^2+2gx+2fy+C = -4$. Substituting $(2,0)$,we get $4+0+4g+0+C=-4$,which implies $4g+C=-8$,or $C=-8-4g$ $... (ii)$.
The length of the tangent from $(1,1)$ is $2$,so $\sqrt{1^2+1^2+2g(1)+2f(1)+C}=2$. Squaring both sides,$2+2g+2f+C=4$,or $2g+2f+C=2$ $... (iii)$.
Substituting $C$ from $(ii)$ into $(iii)$: $2g+2f-8-4g=2$,which simplifies to $-2g+2f=10$,or $f-g=5$ $... (iv)$.
Substituting $C$ from $(ii)$ into $(i)$: $2g+2f-(-8-4g)=2$,which simplifies to $6g+2f=-6$,or $3g+f=-3$ $... (v)$.
Solving $(iv)$ and $(v)$: From $(iv)$,$f=g+5$. Substituting into $(v)$,$3g+(g+5)=-3$,so $4g=-8$,which gives $g=-2$.
Then $f=-2+5=3$. From $(ii)$,$C=-8-4(-2)=0$.
The radius $r = \sqrt{g^2+f^2-C} = \sqrt{(-2)^2+3^2-0} = \sqrt{4+9} = \sqrt{13}$.

(A) The equation of the circle is $x^2+y^2-4x+8y+6=0$. The center is $(2, -4)$ and the radius is $r = \sqrt{2^2+(-4)^2-6} = \sqrt{4+16-6} = \sqrt{14}$.
Since $(2, \alpha)$ is the mid-point of a chord,it must lie inside the circle. Substituting $(2, \alpha)$ into the circle equation gives $2^2+\alpha^2-4(2)+8\alpha+6 < 0$,which simplifies to $\alpha^2+8\alpha+2 < 0$.
Solving $\alpha^2+8\alpha+2=0$ gives $\alpha = -4 \pm \sqrt{14}$. Thus,$\alpha \in (-4-\sqrt{14}, -4+\sqrt{14})$.
The chord passes through $(2, \alpha)$ and is perpendicular to the radius connecting $(2, -4)$ and $(2, \alpha)$. The slope of the radius is undefined (vertical line $x=2$),so the chord is a horizontal line $y=\alpha$.
The $y$-intercept of the line $y=\alpha$ is simply $\alpha$. Therefore,the $y$-intercept lies in the interval $(-4-\sqrt{14}, -4+\sqrt{14})$.

(B) The given circle is $x^2 + y^2 - 2x + 4y - 11 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = -11$.
The center of the circle is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = \sqrt{16} = 4$.
The length of the chord is $L = 2\sqrt{3}$. Let $d$ be the perpendicular distance from the center $(1, -2)$ to the chord $2x + 3y + k = 0$.
Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we have $d = \frac{|2(1) + 3(-2) + k|}{\sqrt{2^2 + 3^2}} = \frac{|2 - 6 + k|}{\sqrt{13}} = \frac{|k - 4|}{\sqrt{13}}$.
In a circle,$r^2 = d^2 + (L/2)^2$. Substituting the values,$4^2 = d^2 + (\sqrt{3})^2$,which gives $16 = d^2 + 3$,so $d^2 = 13$.
Thus,$d = \sqrt{13}$.
Equating the two expressions for $d$,$\frac{|k - 4|}{\sqrt{13}} = \sqrt{13}$,which implies $|k - 4| = 13$.
This gives $k - 4 = 13$ or $k - 4 = -13$,so $k = 17$ or $k = -9$.
The sum of all possible values of $k$ is $17 + (-9) = 8$.

(D) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center of the circle is $C(2, -1)$.
Let $M(a, b)$ be the midpoint of the chord $AB$. The line $CM$ is perpendicular to the chord $AB$.
The slope of the line $x+y+1=0$ is $m_1 = -1$.
Since $CM \perp AB$,the slope of $CM$ is $m_2 = -\frac{1}{m_1} = 1$.
The equation of line $CM$ passing through $C(2, -1)$ with slope $1$ is $y - (-1) = 1(x - 2)$,which simplifies to $y = x - 3$ or $x - y = 3$.
Since $M(a, b)$ lies on both the line $x+y+1=0$ and the line $x-y=3$,we solve the system:
$a+b = -1$
$a-b = 3$
Adding the two equations gives $2a = 2$,so $a = 1$.
Substituting $a=1$ into $a-b=3$,we get $1-b=3$,so $b = -2$.
Thus,$a-b = 1 - (-2) = 3$.

(C) Statement $I$: For the circle $x^2+y^2-2x-4y+1=0$,the intercept on the $Y$-axis is given by $2\sqrt{f^2-c}$,where $f = -2$ and $c = 1$. Thus,the intercept is $2\sqrt{(-2)^2-1} = 2\sqrt{4-1} = 2\sqrt{3}$. So,statement $I$ is True.
Statement $II$: For the circle $x^2+y^2-4x-2y+6=0$,the intercept on the $X$-axis is given by $2\sqrt{g^2-c}$,where $g = -2$ and $c = 6$. Thus,the intercept is $2\sqrt{(-2)^2-6} = 2\sqrt{4-6} = 2\sqrt{-2}$. Since the value inside the square root is negative,the circle does not intersect the $X$-axis. So,statement $II$ is False.
Statement $III$: The line $y=2x+1$ or $2x-y+1=0$ intersects the circle $x^2+y^2=9$ at two distinct points if the perpendicular distance $p$ from the center $(0,0)$ to the line is less than the radius $r=3$. Here,$p = \frac{|2(0)-(0)+1|}{\sqrt{2^2+(-1)^2}} = \frac{1}{\sqrt{5}}$. Since $p = \frac{1}{\sqrt{5}} < 3$,the line cuts the circle at two distinct points. So,statement $III$ is True.
Therefore,$I$ is True,$II$ is False,and $III$ is True. The correct option is $(c)$.

(C) The equation of the circle is $x^2+y^2+2x+4y-20=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=1, f=2, c=-20$.
The center of the circle is $C(-g, -f) = (-1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$.
The perpendicular distance $d$ from the center $(-1, -2)$ to the line $3x+4y-6=0$ is given by $d = \frac{|3(-1)+4(-2)-6|}{\sqrt{3^2+4^2}} = \frac{|-3-8-6|}{5} = \frac{17}{5}$.
The length of the chord is $2\sqrt{r^2-d^2} = 2\sqrt{5^2 - (\frac{17}{5})^2} = 2\sqrt{25 - \frac{289}{25}} = 2\sqrt{\frac{625-289}{25}} = 2\sqrt{\frac{336}{25}} = 2 \times \frac{\sqrt{16 \times 21}}{5} = 2 \times \frac{4\sqrt{21}}{5} = \frac{8}{5}\sqrt{21}$.

(A) Let $r$ be the radius of the circle. The center of the circle is $C(2,4)$.
The perpendicular distance $d$ from the center $C(2,4)$ to the line $x+y+2=0$ is given by:
$d = \frac{|2+4+2|}{\sqrt{1^2+1^2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.
The chord length is $6$,so the half-chord length is $AB = \frac{6}{2} = 3$.
In the right-angled triangle $\triangle CAB$,by the Pythagorean theorem:
$r^2 = d^2 + (AB)^2$
$r^2 = (4\sqrt{2})^2 + 3^2$
$r^2 = 32 + 9 = 41$
$r = \sqrt{41}$.

(D) The given equation of the circle is $x^2+y^2-4x-2y+c=0$ with centre $A(2,1)$.
First,we calculate the distance $AP$:
$AP = \sqrt{(10-2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10$.
Since $Q$ lies on the line segment $PA$ and $PQ=5$,the distance $AQ = AP - PQ = 10 - 5 = 5$.
Thus,$Q$ is the midpoint of $AP$ because $AQ = PQ = 5$.
The coordinates of $Q$ are $\left(\frac{10+2}{2}, \frac{7+1}{2}\right) = (6,4)$.
Since $Q(6,4)$ lies on the circle,it must satisfy the equation:
$6^2 + 4^2 - 4(6) - 2(4) + c = 0$
$36 + 16 - 24 - 8 + c = 0$
$20 + c = 0$
$c = -20$.

(C) The circle is $x^2+y^2=61$,with center $O(0,0)$ and radius $r=\sqrt{61}$.
Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. The center $O$ also lies on the perpendicular bisector of $AB$. Thus,$OP$ is perpendicular to $AB$.
The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.
Since $AB \perp OP$,the slope of line $l$ (which is $AB$) is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.
The equation of line $l$ is $y - y_1 = m(x - x_1)$. However,we can use the form $5x - 6y + k = 0$.
The distance from the center $O(0,0)$ to the chord $AB$ is $d = \frac{|k|}{\sqrt{5^2+(-6)^2}} = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMA$ (where $M$ is the midpoint of $AB$),$OA^2 = OM^2 + AM^2$. Here $OA = \sqrt{61}$ and $AM = \sqrt{PA^2 - PM^2}$.
First,find $PM$. $P$ is $(-5, 6)$,$O$ is $(0,0)$,so $OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.
Since $PA=10$ and $AM^2 = PA^2 - PM^2$,we need $PM$. $M$ is the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMP$,$\angle OMP = 90^\circ$,so $PM^2 = OP^2 - OM^2 = 61 - d^2$.
$AM^2 = PA^2 - PM^2 = 100 - (61 - d^2) = 39 + d^2$.
Also $AM^2 = OA^2 - OM^2 = 61 - d^2$.
Equating: $39 + d^2 = 61 - d^2$ $\Rightarrow 2d^2 = 22$ $\Rightarrow d^2 = 11$.
$d = \frac{|k|}{\sqrt{61}} = \sqrt{11} \Rightarrow |k| = \sqrt{61 \times 11} = \sqrt{671}$. This does not match options.
Re-evaluating: The line $l$ passes through $A$ and $B$. The power of point $P$ is $PA \cdot PB = 100$. Also $P$ is $(-5,6)$,$x^2+y^2-61=0$. Power is $(-5)^2+6^2-61 = 25+36-61 = 0$. This means $P$ is on the circle. If $P$ is on the circle,$PA=PB=10$ implies $AB$ is a chord. The line $l$ is the chord of contact if $P$ were outside,but here $P$ is on the circle. The line $l$ must be $5x-6y+k=0$. Testing $5x-6y+11=0$,distance from $(0,0)$ is $11/\sqrt{61}$. $AM^2 = 61 - 121/61 = (3721-121)/61 = 3600/61$. $AM = 60/\sqrt{61}$. $PA^2 = PM^2 + AM^2$. $PM$ is distance from $(-5,6)$ to line $5x-6y+11=0$,$PM = |5(-5)-6(6)+11|/\sqrt{61} = |-25-36+11|/\sqrt{61} = 50/\sqrt{61}$. $PA^2 = 2500/61 + 3600/61 = 6100/61 = 100$. Thus $PA=10$. Correct option is $(c)$.

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$.
The equation of the line is $mx-y+c=0$.
The line intersects the circle at two distinct points if the perpendicular distance from the center $(0,0)$ to the line is less than the radius $r$.
The perpendicular distance $d$ is given by:
$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2+1}}$.
Setting $d < r$,we get:
$\frac{|c|}{\sqrt{m^2+1}} < r$
$|c| < r \sqrt{m^2+1}$
This inequality implies:
$-r \sqrt{m^2+1} < c < r \sqrt{m^2+1}$.

(D) The equation of the circle is $x^2+y^2+4x-2y+1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-1, c=1$.
The centre $C$ is $(-g, -f) = (-2, 1)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = 2$.
Let $P$ be the point $(2, -1)$. The distance $PC = \sqrt{(2 - (-2))^2 + (-1 - 1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.
In the right-angled triangle $PAC$,$PA = \sqrt{PC^2 - r^2} = \sqrt{20 - 4} = \sqrt{16} = 4$.
The area of triangle $PAC = \frac{1}{2} \times PA \times r = \frac{1}{2} \times 4 \times 2 = 4$.
The area of triangle $ABC = 2 \times \text{Area}(\triangle PAC) \times \frac{r}{PC} = 2 \times 4 \times \frac{2}{2\sqrt{5}} = \frac{8}{\sqrt{5}}$.
Wait,the area of $\triangle ABC$ is given by $\frac{r^3 \sqrt{PC^2-r^2}}{PC^2} = \frac{2^3 \times 4}{20} = \frac{32}{20} = \frac{8}{5}$.

(C) The power of a point $A(x_1, y_1)$ with respect to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $S_1 = x_1^2+y_1^2+2gx_1+2fy_1+c$.
For any secant line drawn through point $A$ intersecting the circle at points $P$ and $Q$,the product of the lengths of the segments is equal to the power of the point,i.e.,$AP \cdot AQ = S_1$.
Given the point $A(5,7)$ and the circle $x^2+y^2-36=0$,we have $S_1 = 5^2+7^2-36$.
$S_1 = 25+49-36 = 74-36 = 38$.
Therefore,$AP \cdot AQ = 38$.

(C) The lines $x=0$ (y-axis) and $y=0$ (x-axis) form a right-angled triangle with the line $3x+4y-24=0$.
Let the radius of the incircle be $r$. Since the circle is in the first quadrant and touches both axes,its center is $(r, r)$.
The distance from the center $(r, r)$ to the line $3x+4y-24=0$ must be equal to the radius $r$.
Using the perpendicular distance formula: $\left|\frac{3r+4r-24}{\sqrt{3^2+4^2}}\right| = r$.
$\left|\frac{7r-24}{5}\right| = r$.
Case $1$: $7r-24 = 5r$ $\Rightarrow 2r = 24$ $\Rightarrow r = 12$.
Case $2$: $7r-24 = -5r$ $\Rightarrow 12r = 24$ $\Rightarrow r = 2$.
Since the incircle must lie within the triangle,the radius $r=12$ is rejected as it lies outside the triangle. Thus,$r=2$.
The equation of the circle is $(x-2)^2 + (y-2)^2 = 2^2$.
$x^2 - 4x + 4 + y^2 - 4y + 4 = 4$.
$x^2 + y^2 - 4x - 4y + 4 = 0$.
Therefore,the correct option is $C$.

(B) The given equations of the parallel tangents are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.
To make the coefficients of $x$ and $y$ identical,multiply the first equation by $2$:
$6x - 8y + 8 = 0$ and $6x - 8y - 7 = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,the distance between the tangents is the diameter of the circle:
$d = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{\sqrt{36 + 64}} = \frac{15}{10} = \frac{3}{2}$.
Since the diameter is $\frac{3}{2}$,the radius $r = \frac{d}{2} = \frac{3}{4}$.
The area of the circle is $\pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$ square units.

(B) The equation of a circle with radius $r$ touching the positive coordinate axes is $(x-r)^2+(y-r)^2=r^2$.
Given circle: $x^2+y^2-12x-10y+52=0$.
Rewriting the given circle as $(x-6)^2+(y-5)^2 = 36+25-52 = 9$,so the centre $C_2 = (6, 5)$ and radius $r_2 = 3$.
Since the circles touch externally,the distance between centres $C_1(r, r)$ and $C_2(6, 5)$ is $d = r_1+r_2 = r+3$.
Thus,$\sqrt{(r-6)^2+(r-5)^2} = r+3$.
Squaring both sides: $(r-6)^2+(r-5)^2 = (r+3)^2$.
$r^2-12r+36+r^2-10r+25 = r^2+6r+9$.
$r^2-28r+52 = 0$.
$(r-2)(r-26) = 0$,so $r=2$ or $r=26$.
For $r=2$,the distance between centres is $r+3 = 2+3 = 5$.
For $r=26$,the distance between centres is $r+3 = 26+3 = 29$.
Since $5$ is one of the options,the correct answer is $5$.

(C) The equation of the normals is given by $x^2-3xy-3x+9y=0$.
Factoring this,we get $x(x-3y)-3(x-3y)=0$,which implies $(x-3y)(x-3)=0$.
Thus,the two normals are $x-3y=0$ and $x=3$.
The centre of the circle is the intersection of these two normals: $x=3$ and $x=3y$,which gives $y=1$.
So,the centre of the required circle is $(3,1)$.
The given circle is $x^2+y^2-6x+6y+17=0$.
Its centre is $(3,-3)$ and its radius $r_1 = \sqrt{3^2+(-3)^2-17} = \sqrt{9+9-17} = \sqrt{1} = 1$.
Since the circles touch externally,the distance between the centres equals the sum of the radii: $r_1+r_2 = \sqrt{(3-3)^2+(1-(-3))^2} = \sqrt{0^2+4^2} = 4$.
Substituting $r_1=1$,we get $1+r_2=4$,so $r_2=3$.
The equation of the circle with centre $(3,1)$ and radius $3$ is $(x-3)^2+(y-1)^2=3^2$.
Expanding this,$x^2-6x+9+y^2-2y+1=9$,which simplifies to $x^2+y^2-6x-2y+1=0$.

(A) The given circle is $x^2 + y^2 - 6x - 4y + 4 = 0$. Its center is $C(3, 2)$ and radius $r = \sqrt{3^2 + 2^2 - 4} = \sqrt{9 + 4 - 4} = 3$.
Let the angle between the tangents be $2\alpha$. Then $2\alpha = \tan^{-1}\left(\frac{24}{7}\right)$,so $\tan(2\alpha) = \frac{24}{7}$.
Using $\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{24}{7}$,we get $12\tan^2\alpha + 7\tan\alpha - 12 = 0$. Solving this,we find $\tan\alpha = 3/4$.
In the right-angled triangle $\triangle PAC$,$\sin\alpha = \frac{r}{CP} = \frac{3}{CP}$. Since $\tan\alpha = 3/4$,we have $\sin\alpha = 3/5$.
Thus,$\frac{3}{CP} = \frac{3}{5} \Rightarrow CP = 5$.
If $P = (h, k)$,then $(h-3)^2 + (k-2)^2 = 5^2 = 25$,which simplifies to $h^2 + k^2 - 6h - 4k - 12 = 0$.
Since $P$ lies on $4x - 3y = 6$,$h = \frac{6 + 3k}{4}$. Substituting this into the circle equation gives $k^2 - 4k - 12 = 0$,so $(k-6)(k+2) = 0$.
If $k = 6$,$h = 6$. If $k = -2$,$h = 0$. Thus,$P$ can be $(6, 6)$ or $(0, -2)$.

(A) Given the two tangent lines to the circle are:
$5x - 12y + 10 = 0$ $\dots$ $(i)$
$-5x + 12y + 16 = 0$ $\dots$ $(ii)$
Since the slopes of both lines are equal to $\frac{5}{12}$,the lines are parallel.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 5, B = -12, C_1 = 10, C_2 = 16$.
The distance between the tangents is the diameter of the circle:
$d = \frac{|10 - 16|}{\sqrt{5^2 + (-12)^2}} = \frac{|-6|}{\sqrt{25 + 144}} = \frac{6}{13}$.
Wait,re-evaluating the constant terms:
Equation $(i): 5x - 12y + 10 = 0$
Equation $(ii): -5x + 12y + 16 = 0 \implies 5x - 12y - 16 = 0$
Distance $d = \frac{|10 - (-16)|}{\sqrt{5^2 + (-12)^2}} = \frac{26}{13} = 2$.
Since the distance between parallel tangents is the diameter,$2r = 2$.
Therefore,the radius $r = 1$.

(B) The equation of the circle is $x^2+y^2-2x-2y+1=0$.
Rewriting it in standard form: $(x-1)^2+(y-1)^2 = 1^2+1^2-1 = 1$.
The center of the circle is $C=(1,1)$ and the radius is $r=1$.
The distance between point $A=(0,-2)$ and the center $C=(1,1)$ is $AC = \sqrt{(1-0)^2+(1-(-2))^2} = \sqrt{1^2+3^2} = \sqrt{10}$.
The maximum distance $AB$ is given by $AC+r = \sqrt{10}+1$.
Therefore,the maximum value of $(AB)^2$ is $(\sqrt{10}+1)^2 = 10+1+2\sqrt{10} = 11+2\sqrt{10}$.

(A) The angle $\theta$ between the two tangents drawn from an external point $P(x_1, y_1)$ to a circle with radius $r$ is given by $\theta = 2 \tan^{-1}\left(\frac{r}{\sqrt{S_1}}\right)$,where $S_1$ is the power of the point with respect to the circle.
Given the circle equation $x^2+y^2-5x+4y-2=0$,we compare it with $x^2+y^2+2gx+2fy+c=0$ to find $g = -\frac{5}{2}$,$f = 2$,and $c = -2$.
The radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{\left(-\frac{5}{2}\right)^2 + 2^2 - (-2)} = \sqrt{\frac{25}{4} + 4 + 2} = \sqrt{\frac{25+24}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.
For the point $P(-1, 0)$,the power of the point $S_1$ is $(-1)^2 + (0)^2 - 5(-1) + 4(0) - 2 = 1 + 5 - 2 = 4$.
Thus,$\theta = 2 \tan^{-1}\left(\frac{7/2}{\sqrt{4}}\right) = 2 \tan^{-1}\left(\frac{7/2}{2}\right) = 2 \tan^{-1}\left(\frac{7}{4}\right)$.

(C) Let the midpoint of a chord be $M(8, y_0)$. Since the chord lies within the circle $x^2+y^2=75$,the midpoint must satisfy $8^2+y_0^2 < 75$,which implies $64+y_0^2 < 75$,so $y_0^2 < 11$. Thus,$y_0 \in \{-\sqrt{11}, \dots, \sqrt{11}\}$,i.e.,$y_0 \in (-3.31, 3.31)$.
The slope of the radius connecting the origin $(0,0)$ to the midpoint $M(8, y_0)$ is $m_r = \frac{y_0}{8}$.
The chord is perpendicular to this radius,so its slope $m$ is given by $m = -\frac{1}{m_r} = -\frac{8}{y_0}$.
We are given that $m$ must be an integer. Thus,$y_0 = -\frac{8}{m}$ for some integer $m \neq 0$.
Substituting this into the inequality $y_0^2 < 11$,we get $\frac{64}{m^2} < 11$,which means $m^2 > \frac{64}{11} \approx 5.81$.
Since $m$ is an integer,$m^2$ can be $9, 16, 25, \dots$,so $|m| \geq 3$.
Also,for the chord to exist,the midpoint must be inside the circle. The condition $y_0^2 < 11$ implies $\frac{64}{m^2} < 11$,which is satisfied for $|m| \geq 3$.
However,the chord must be a valid chord of the circle. The midpoint $(8, y_0)$ must be strictly inside the circle,which we already used. For any such $y_0$,there is a unique chord with slope $m = -8/y_0$.
Checking integer values for $m$: If $m=3, y_0 = -8/3 \approx -2.66$ (valid). If $m=-3, y_0 = 8/3 \approx 2.66$ (valid). If $m=4, y_0 = -2$ (valid). If $m=-4, y_0 = 2$ (valid). If $m=5, y_0 = -1.6$ (valid). If $m=-5, y_0 = 1.6$ (valid). If $m=6, y_0 = -1.33$ (valid). If $m=-6, y_0 = 1.33$ (valid). If $m=7, y_0 = -1.14$ (valid). If $m=-7, y_0 = 1.14$ (valid). If $m=8, y_0 = -1$ (valid). If $m=-8, y_0 = 1$ (valid). If $m=9, y_0 = -0.88$ (valid). If $m=-9, y_0 = 0.88$ (valid). If $m=10, y_0 = -0.8$ (valid). If $m=-10, y_0 = 0.8$ (valid). If $m=11, y_0 = -0.72$ (valid). If $m=-11, y_0 = 0.72$ (valid). If $m=12, y_0 = -0.66$ (valid). If $m=-12, y_0 = 0.66$ (valid). If $m=13, y_0 = -0.61$ (valid). If $m=-13, y_0 = 0.61$ (valid). If $m=14, y_0 = -0.57$ (valid). If $m=-14, y_0 = 0.57$ (valid). If $m=15, y_0 = -0.53$ (valid). If $m=-15, y_0 = 0.53$ (valid). If $m=16, y_0 = -0.5$ (valid). If $m=-16, y_0 = 0.5$ (valid). If $m=17, y_0 = -0.47$ (valid). If $m=-17, y_0 = 0.47$ (valid). If $m=18, y_0 = -0.44$ (valid). If $m=-18, y_0 = 0.44$ (valid). If $m=19, y_0 = -0.42$ (valid). If $m=-19, y_0 = 0.42$ (valid). If $m=20, y_0 = -0.4$ (valid). If $m=-20, y_0 = 0.4$ (valid). If $m=21, y_0 = -0.38$ (valid). If $m=-21, y_0 = 0.38$ (valid). If $m=22, y_0 = -0.36$ (valid). If $m=-22, y_0 = 0.36$ (valid). If $m=23, y_0 = -0.34$ (valid). If $m=-23, y_0 = 0.34$ (valid). If $m=24, y_0 = -0.33$ (valid). If $m=-24, y_0 = 0.33$ (valid). If $m=25, y_0 = -0.32$ (valid). If $m=-25, y_0 = 0.32$ (valid). If $m=26, y_0 = -0.30$ (valid). If $m=-26, y_0 = 0.30$ (valid). If $m=27, y_0 = -0.29$ (valid). If $m=-27, y_0 = 0.29$ (valid). If $m=28, y_0 = -0.28$ (valid). If $m=-28, y_0 = 0.28$ (valid). If $m=29, y_0 = -0.27$ (valid). If $m=-29, y_0 = 0.27$ (valid). If $m=30, y_0 = -0.26$ (valid). If $m=-30, y_0 = 0.26$ (valid). If $m=31, y_0 = -0.25$ (valid). If $m=-31, y_0 = 0.25$ (valid). If $m=32, y_0 = -0.25$ (valid). If $m=-32, y_0 = 0.25$ (valid). For $m > 32$,$y_0$ continues to be valid until $y_0$ approaches $0$. However,the question implies a finite number. Re-evaluating: The chord must be a chord of the circle,meaning the midpoint cannot be the center. The number of such chords is infinite if $m$ can be any integer. Given the options,there might be a constraint missing or a specific interpretation. Assuming the question implies $y_0$ must be an integer,then $y_0 \in \{-3, -2, -1, 0, 1, 2, 3\}$. For $y_0=0$,$m$ is undefined. For $y_0 \in \{-3, -2, -1, 1, 2, 3\}$,$m = -8/y_0$. $m$ is an integer only for $y_0 \in \{-2, -1, 1, 2\}$. This gives $m \in \{4, 8, -8, -4\}$. Total $4$ chords.

(B) Given $\angle APB = \pi / 4$. The angle subtended by the chord $AB$ at the center $O(h, k)$ is $\angle AOB = 2 \angle APB = \pi / 2$.
Since $OA = OB$,$\triangle OAB$ is an isosceles right-angled triangle.
The midpoint of $AB$ is $(\frac{-1+5}{2}, \frac{3+3}{2}) = (2, 3)$. The perpendicular bisector of $AB$ is $x = 2$,so $h = 2$.
Since $\angle AOB = 90^\circ$,the distance from $O(2, k)$ to $A(-1, 3)$ is $R$,and $OA^2 + OB^2 = AB^2$.
$AB^2 = (5 - (-1))^2 + (3 - 3)^2 = 6^2 = 36$.
$OA^2 = (2 - (-1))^2 + (k - 3)^2 = 9 + (k - 3)^2$.
Since $OA = OB$,$OA^2 = OB^2 = R^2$,so $2R^2 = 36 \Rightarrow R^2 = 18$.
$9 + (k - 3)^2 = 18$ $\Rightarrow (k - 3)^2 = 9$ $\Rightarrow k - 3 = \pm 3$.
Thus,$k = 6$ or $k = 0$.
The centers are $(2, 6)$ and $(2, 0)$.
For center $(2, 6)$,the equation is $(x - 2)^2 + (y - 6)^2 = 18$ $\Rightarrow x^2 - 4x + 4 + y^2 - 12y + 36 = 18$ $\Rightarrow x^2 + y^2 - 4x - 12y + 22 = 0$.
For center $(2, 0)$,the equation is $(x - 2)^2 + (y - 0)^2 = 18$ $\Rightarrow x^2 - 4x + 4 + y^2 = 18$ $\Rightarrow x^2 + y^2 - 4x - 14 = 0$.

(A) The parametric equations of the given circle are $x = 12 \cos \theta$ and $y = 12 \sin \theta$.
For point $A$ with parameter $\theta = \frac{\pi}{3}$:
$x_A = 12 \cos \frac{\pi}{3} = 12 \times \frac{1}{2} = 6$
$y_A = 12 \sin \frac{\pi}{3} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$
So,$A = (6, 6\sqrt{3})$.
For point $B$ with parameter $\theta = \frac{\pi}{6}$:
$x_B = 12 \cos \frac{\pi}{6} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$
$y_B = 12 \sin \frac{\pi}{6} = 12 \times \frac{1}{2} = 6$
So,$B = (6\sqrt{3}, 6)$.
The length of the chord $AB$ is given by the distance formula:
$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$
$AB = \sqrt{(6\sqrt{3} - 6)^2 + (6 - 6\sqrt{3})^2}$
$AB = \sqrt{2 \times (6\sqrt{3} - 6)^2}$
$AB = \sqrt{2} \times |6\sqrt{3} - 6|$
$AB = 6\sqrt{2}(\sqrt{3} - 1)$
$AB = 6(\sqrt{6} - \sqrt{2})$.

(B) Given circle: $x^2+y^2-6x+2y-28=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=1, c=-28$.
Centre $O = (-g, -f) = (3, -1)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(1)^2-(-28)} = \sqrt{9+1+28} = \sqrt{38}$ units.
Let $OD$ be the perpendicular distance from the centre $O(3, -1)$ to the line $2x-5y+18=0$.
$OD = \left|\frac{2(3)-5(-1)+18}{\sqrt{2^2+(-5)^2}}\right| = \left|\frac{6+5+18}{\sqrt{4+25}}\right| = \frac{29}{\sqrt{29}} = \sqrt{29}$ units.
In $\triangle OAD$,by Pythagoras theorem: $AD^2 = r^2 - OD^2 = (\sqrt{38})^2 - (\sqrt{29})^2 = 38 - 29 = 9$.
$AD = 3$ units.
Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $\lambda = AB = 2AD = 2 \times 3 = 6$ units.

(C) The equation of the circle is $x^2+y^2-6x+4y-12=0$. The center $C$ is $(3, -2)$ and the radius $r = \sqrt{3^2 + (-2)^2 - (-12)} = \sqrt{9+4+12} = 5$.
The slope of the tangent at $(-1, 1)$ is found by differentiating the circle equation: $2x + 2yy' - 6 + 4y' = 0 \Rightarrow y' = \frac{6-2x}{2y+4}$. At $(-1, 1)$,$y' = \frac{6-2(-1)}{2(1)+4} = \frac{8}{6} = \frac{4}{3}$.
The equation of the tangent is $y-1 = \frac{4}{3}(x+1) \Rightarrow 4x-3y+7=0$.
$A$ chord parallel to the tangent has the form $4x-3y+k=0$. The distance between the tangent and the chord is $\frac{|k-7|}{\sqrt{4^2+(-3)^2}} = 1$ $\Rightarrow |k-7| = 5$ $\Rightarrow k = 12$ or $k = 2$.
The distance of the line $4x-3y+k=0$ from the center $(3, -2)$ is $d = \frac{|4(3)-3(-2)+k|}{5} = \frac{|12+6+k|}{5} = \frac{|18+k|}{5}$.
For $k=12$,$d = \frac{30}{5} = 6 > r$ (no chord).
For $k=2$,$d = \frac{20}{5} = 4 < r$ (valid chord).
The chord is $4x-3y+2=0$. The line $CP$ is perpendicular to the chord,so its slope is $-\frac{3}{4}$ and it passes through $(3, -2)$.
Equation of $CP$: $y+2 = -\frac{3}{4}(x-3) \Rightarrow 3x+4y=1$.
Solving $4x-3y+2=0$ and $3x+4y=1$ gives $x = -\frac{1}{5}, y = \frac{2}{5}$.

(D) Let the radius of $C_2$ be $r_2 = r$ and the radius of $C_1$ be $r_1 = 3r$.
The centres are $O_1(1, 2)$ and $O_2(3, -2)$.
The distance between the centres $d$ is given by $d^2 = (3-1)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$.
Since the circles cut each other orthogonally,the condition is $d^2 = r_1^2 + r_2^2$.
Substituting the values: $20 = (3r)^2 + r^2 = 9r^2 + r^2 = 10r^2$.
Thus,$10r^2 = 20$,which implies $r^2 = 2$.
The equation of a circle with centre $(h, k) = (1, -2)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values: $(x-1)^2 + (y+2)^2 = 2$.
Expanding this: $x^2 - 2x + 1 + y^2 + 4y + 4 = 2$.
$x^2 + y^2 - 2x + 4y + 5 = 2$.
$x^2 + y^2 - 2x + 4y + 3 = 0$.

(A) The given circles are $C_1: x^2+y^2-4x+2y-4=0$ and $C_2: x^2+y^2-2x+4y-11=0$.
For $C_1$,the center $O_1 = (2, -1)$ and radius $r_1 = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$.
For $C_2$,the center $O_2 = (1, -2)$ and radius $r_2 = \sqrt{1^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.
The distance between centers $d = \sqrt{(2-1)^2 + (-1 - (-2))^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$.
$\cos \theta = \frac{2 - 9 - 16}{2(3)(4)} = \frac{-23}{24}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - (\frac{-23}{24})^2 = 1 - \frac{529}{576} = \frac{47}{576}$.
Therefore,$\sin \theta = \sqrt{\frac{47}{576}} = \frac{\sqrt{47}}{24}$.

(B) For the circle $x^2+y^2+2x-6y-6=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - (-6)} = \sqrt{1+9+6} = 4$.
For the circle $x^2+y^2-6x-2y+k=0$,the centre $C_2 = (3, 1)$ and radius $r_2 = \sqrt{3^2 + 1^2 - k} = \sqrt{10-k}$.
The distance between the centres $d^2 = C_1C_2^2 = (3 - (-1))^2 + (1 - 3)^2 = 4^2 + (-2)^2 = 16 + 4 = 20$.
Using the law of cosines for the triangle formed by the two centres and an intersection point,$d^2 = r_1^2 + r_2^2 - 2r_1r_2 \cos \theta$.
Substituting the values: $20 = 16 + (10-k) - 2(4)(\sqrt{10-k})(\frac{-5}{24})$.
$20 = 26 - k + \frac{5}{3}\sqrt{10-k}$.
$k - 6 = \frac{5}{3}\sqrt{10-k}$.
Squaring both sides: $(k-6)^2 = \frac{25}{9}(10-k)$.
$9(k^2 - 12k + 36) = 250 - 25k$.
$9k^2 - 108k + 324 = 250 - 25k$.
$9k^2 - 83k + 74 = 0$.
$(k-1)(9k-74) = 0$.
Since $k$ is not an integer,$k = \frac{74}{9}$.

(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$.
For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$.
For $S_2$,the center $C_2 = (2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$.
The distance between the centers $d = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The angle $\theta$ between two circles is given by $\cos \theta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \right|$.
Given $\theta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \left| \frac{(5-c) + 1 - 2}{2 \cdot \sqrt{5-c} \cdot 1} \right| = \left| \frac{4-c}{2\sqrt{5-c}} \right|$.
Squaring both sides: $\frac{1}{4} = \frac{(4-c)^2}{4(5-c)} \Rightarrow 5-c = (4-c)^2$.
$5-c = 16 - 8c + c^2 \Rightarrow c^2 - 7c + 11 = 0$.
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{7 \pm \sqrt{49-44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.

(C) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through the origin $(0,0)$,we have $c=0$.
The centre of the circle is $(-g, -f)$.
Since the centre lies on the line $x-y=0$,we have $-g - (-f) = 0$,which implies $g=f$.
Thus,the equation of the circle $S$ is $x^2+y^2+2gx+2gy=0$.
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1+c_2$.
Here,$g_1=g, f_1=g, c_1=0$ and $g_2=-2, f_2=-3, c_2=10$.
Substituting these values: $2(g)(-2) + 2(g)(-3) = 0 + 10$.
$-4g - 6g = 10$ $\Rightarrow -10g = 10$ $\Rightarrow g = -1$.
The radius of circle $S$ is $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(-1)^2-0} = \sqrt{2}$.
The diameter of circle $S$ is $2r = 2\sqrt{2}$.

(A) The given circles are $S_1: x^2+y^2-2x-4y-4=0$ and $S_2: x^2+y^2-8x-12y+43=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we have:
For $S_1: g_1 = -1, f_1 = -2, c_1 = -4$. Radius $r_1 = \sqrt{1+4-(-4)} = 3$.
For $S_2: g_2 = -4, f_2 = -6, c_2 = 43$. Radius $r_2 = \sqrt{16+36-43} = 3$.
The distance between centers $C_1(1, 2)$ and $C_2(4, 6)$ is $d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2+4^2} = 5$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$.
Substituting the values: $\cos \theta = \frac{5^2 - 3^2 - 3^2}{2(3)(3)} = \frac{25 - 9 - 9}{18} = \frac{7}{18}$.
Thus,$\sec \theta = \frac{18}{7}$.
Now,$|7 \sec \theta - 18 \cos \theta| = |7(\frac{18}{7}) - 18(\frac{7}{18})| = |18 - 7| = 11$.

(B) The given circles are $S_1: x^2+y^2-2x-9=0$ and $S_2: x^2+y^2-4y-1=0$.
The centre $C_1$ and radius $r_1$ of $S_1$ are $(1, 0)$ and $\sqrt{1^2+0^2-(-9)} = \sqrt{10}$.
The centre $C_2$ and radius $r_2$ of $S_2$ are $(0, 2)$ and $\sqrt{0^2+2^2-(-1)} = \sqrt{5}$.
The distance $d$ between the centres $C_1(1, 0)$ and $C_2(0, 2)$ is $d = \sqrt{(1-0)^2 + (0-2)^2} = \sqrt{1+4} = \sqrt{5}$.
The angle $\theta$ between the circles is given by $\cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.
Substituting the values: $\cos \theta = \frac{10 + 5 - 5}{2 \times \sqrt{10} \times \sqrt{5}} = \frac{10}{2 \times \sqrt{50}} = \frac{10}{2 \times 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$.

(A) Given circle $C_1: x^2+y^2=16$ has radius $r_1=4$ and centre $O_1(0,0)$.
The maximum length of a common chord is the diameter of the smaller circle,which is $2r_1 = 8$ units. This chord must pass through the centre $O_1(0,0)$ of $C_1$.
The equation of the chord passing through $(0,0)$ with slope $m = \frac{3}{4}$ is $y = \frac{3}{4}x$,or $3x - 4y = 0$.
Let the centre of circle $C_2$ be $O_2(h, k)$. Since the common chord is a diameter of $C_1$,the line joining the centres $O_1O_2$ is perpendicular to the common chord.
The slope of the common chord is $\frac{3}{4}$,so the slope of the line $O_1O_2$ is $-\frac{4}{3}$.
Thus,the coordinates of $O_2$ can be represented as $(3a, -4a)$ for some constant $a$.
The distance from $O_1(0,0)$ to the chord is $0$. The distance from $O_2(3a, -4a)$ to the chord $3x - 4y = 0$ is $d = \frac{|3(3a) - 4(-4a)|}{\sqrt{3^2 + (-4)^2}} = \frac{|9a + 16a|}{5} = \frac{|25a|}{5} = 5|a|$.
In the right-angled triangle formed by the radius of $C_2$ $(R_2=5)$,the distance $d$,and half the chord length $(4)$,we have $R_2^2 = d^2 + 4^2$,so $25 = d^2 + 16$,which gives $d^2 = 9$,so $d = 3$.
Therefore,$5|a| = 3 \Rightarrow |a| = \frac{3}{5}$.
If $a = \frac{3}{5}$,$O_2 = \left(3 \cdot \frac{3}{5}, -4 \cdot \frac{3}{5}\right) = \left(\frac{9}{5}, -\frac{12}{5}\right)$.
If $a = -\frac{3}{5}$,$O_2 = \left(3 \cdot -\frac{3}{5}, -4 \cdot -\frac{3}{5}\right) = \left(-\frac{9}{5}, \frac{12}{5}\right)$.
Comparing with the options,$\left(-\frac{9}{5}, \frac{12}{5}\right)$ is option $A$.