Geometrical problems regarding circle and its properties Questions in English

(D) The equation of the circle is $x^2 + y^2 + 10x + 12y + c = 0$.
The center is $(-5, -6)$ and the radius $R = \sqrt{(-5)^2 + (-6)^2 - c} = \sqrt{25 + 36 - c} = \sqrt{61 - c}$.
For an equilateral triangle inscribed in a circle of radius $R$,the side length $a$ is given by $a = R\sqrt{3}$.
The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (R\sqrt{3})^2 = \frac{3\sqrt{3}}{4} R^2$.
Given the area is $27\sqrt{3}$,we have $\frac{3\sqrt{3}}{4} R^2 = 27\sqrt{3}$.
$R^2 = 27 \times \frac{4}{3} = 36$.
Since $R^2 = 61 - c$,we have $61 - c = 36$.
$c = 61 - 36 = 25$.

(C) The equation of the circle is $x^2 + y^2 - 6x + 8y - 103 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$,$f = 4$,and $c = -103$.
The centre of the circle is $(-g, -f) = (3, -4)$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 4^2 - (-103)} = \sqrt{9 + 16 + 103} = \sqrt{128} = 8\sqrt{2}$.
Since the sides of the square are parallel to the coordinate axes,the vertices are at a distance $r/\sqrt{2} = 8$ from the centre along the horizontal and vertical directions.
The vertices are $(3 \pm 8, -4 \pm 8)$,which are $(11, 4), (11, -12), (-5, 4), (-5, -12)$.
The distances of these vertices from the origin $(0, 0)$ are:
$d_1 = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}$
$d_2 = \sqrt{11^2 + (-12)^2} = \sqrt{121 + 144} = \sqrt{265}$
$d_3 = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$
$d_4 = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
The minimum distance is $\sqrt{41}$.

(B) Let the centres of the two circles be $A$ and $B$. Since the radii are equal,let the radius be $r$. The circles intersect at $D(0, 1)$ and $E(0, -1)$.
Let the centre of the first circle be $A(-h, 0)$ and the second be $B(h, 0)$.
The equation of the circle with centre $A(-h, 0)$ is $(x+h)^2 + y^2 = r^2$. Since it passes through $(0, 1)$,we have $h^2 + 1 = r^2$.
The tangent at $(0, 1)$ to this circle has the equation $x(0+h) + y(1) = r^2$,which simplifies to $hx + y = r^2$.
This tangent passes through the centre of the other circle $B(h, 0)$,so $h(h) + 0 = r^2$,which means $h^2 = r^2$.
Substituting $r^2 = h^2$ into $h^2 + 1 = r^2$ gives $h^2 + 1 = h^2$,which is impossible. Let's re-evaluate.
The tangent at $(0, 1)$ to the circle with centre $A$ is perpendicular to the radius $AD$. The slope of $AD$ is $\frac{1-0}{0-(-h)} = \frac{1}{h}$. Thus,the slope of the tangent is $-h$.
The equation of the tangent is $y - 1 = -h(x - 0)$,or $y = -hx + 1$.
This line passes through the centre of the other circle $B(h, 0)$,so $0 = -h(h) + 1$,which gives $h^2 = 1$,so $h = 1$.
The distance between the centres $A(-1, 0)$ and $B(1, 0)$ is $1 - (-1) = 2$.

(D) The circle is $x^2 + y^2 = 16$,so its radius $r = 4$. The distance $p$ from the center $(0, 0)$ to the line $x + y - n = 0$ is $p = \frac{|0 + 0 - n|}{\sqrt{1^2 + 1^2}} = \frac{n}{\sqrt{2}}$.
For a chord to exist,$p < r$,so $\frac{n}{\sqrt{2}} < 4$,which means $n < 4\sqrt{2} \approx 5.65$. Since $n \in N$,$n \in \{1, 2, 3, 4, 5\}$.
The length of the chord $L$ is given by $L = 2\sqrt{r^2 - p^2} = 2\sqrt{16 - \frac{n^2}{2}} = \sqrt{64 - 2n^2}$.
The square of the length is $L^2 = 64 - 2n^2$.
For $n = 1, L^2 = 64 - 2(1) = 62$.
For $n = 2, L^2 = 64 - 2(4) = 56$.
For $n = 3, L^2 = 64 - 2(9) = 46$.
For $n = 4, L^2 = 64 - 2(16) = 32$.
For $n = 5, L^2 = 64 - 2(25) = 14$.
The sum of the squares is $62 + 56 + 46 + 32 + 14 = 210$.

(B) Let the vertices of the rectangle be $A(-8, 5)$ and $B(6, 5)$. Since $AB$ is a side of the rectangle,the length of $AB = |6 - (-8)| = 14$.
Let the other two vertices be $D(-8, k)$ and $C(6, k)$.
The center of the circle is the midpoint of the diagonal $AC$ (or $BD$).
Midpoint of $AC = \left( \frac{-8 + 6}{2}, \frac{5 + k}{2} \right) = \left( -1, \frac{5 + k}{2} \right)$.
Since the center lies on the diameter $3y = x + 7$,we substitute the coordinates into the equation:
$3\left( \frac{5 + k}{2} \right) = -1 + 7$
$3\left( \frac{5 + k}{2} \right) = 6$
$\frac{5 + k}{2} = 2$
$5 + k = 4$
$k = -1$.
The length of the side $BC = |5 - (-1)| = 6$.
Area of the rectangle = $AB \times BC = 14 \times 6 = 84 \text{ sq. units}$.

(C) Let the center of the circle be $(3, r)$ or $(3, -r)$ and the radius be $r$.
Since the circle touches the $x-$ axis at $(3, 0)$,its equation is $(x - 3)^2 + (y - r)^2 = r^2$ or $(x - 3)^2 + (y + r)^2 = r^2$.
Expanding this,we get $x^2 - 6x + 9 + y^2 \mp 2ry + r^2 = r^2$,which simplifies to $x^2 + y^2 - 6x \mp 2ry + 9 = 0$.
The $y-$ intercept is given by $2\sqrt{f^2 - c}$,where $f = \mp r$ and $c = 9$.
Given the intercept length is $8$,we have $2\sqrt{r^2 - 9} = 8$.
$\sqrt{r^2 - 9} = 4 \implies r^2 - 9 = 16 \implies r^2 = 25 \implies r = 5$.
The equations of the circles are $x^2 + y^2 - 6x + 10y + 9 = 0$ and $x^2 + y^2 - 6x - 10y + 9 = 0$.
Checking the point $(3, 10)$ in the second equation: $(3)^2 + (10)^2 - 6(3) - 10(10) + 9 = 9 + 100 - 18 - 100 + 9 = 0$.
Thus,the circle passes through $(3, 10)$.

(A) Given: Arc length $l = 37.4 \, cm$ and central angle $\theta = 60^{\circ}$.
First, convert the angle from degrees to radians: $\theta = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \, \text{radians}$.
Using the formula for arc length $l = r\theta$, where $r$ is the radius, we have $r = \frac{l}{\theta}$.
Substituting the values: $r = \frac{37.4}{\pi / 3} = \frac{37.4 \times 3}{\pi}$.
Using $\pi = \frac{22}{7}$, we get $r = \frac{37.4 \times 3 \times 7}{22}$.
$r = \frac{112.2 \times 7}{22} = \frac{785.4}{22} = 35.7 \, cm$.

(A) The length of the minute hand is the radius of the circular path,$r = 1.5 \, \text{cm}$.
In $60$ minutes,the minute hand completes one full revolution ($360^\circ$ or $2\pi$ radians).
In $40$ minutes,the fraction of the revolution covered is $\frac{40}{60} = \frac{2}{3}$.
Therefore,the angle $\theta$ subtended at the center is $\theta = \frac{2}{3} \times 2\pi = \frac{4\pi}{3}$ radians.
The distance $l$ moved by the tip is given by the arc length formula $l = r\theta$.
$l = 1.5 \times \frac{4\pi}{3} = 0.5 \times 4\pi = 2\pi \, \text{cm}$.
Using $\pi = 3.14$,we get $l = 2 \times 3.14 = 6.28 \, \text{cm}$.

(A) Diameter of the circle $= 40 \, cm$.
Radius $(r)$ of the circle $= \frac{40}{2} \, cm = 20 \, cm$.
Let $AB$ be a chord of length $20 \, cm$ in the circle with center $O$.
In $\Delta OAB$,$OA = OB = r = 20 \, cm$ and $AB = 20 \, cm$.
Since all sides are equal,$\Delta OAB$ is an equilateral triangle.
Therefore,the central angle $\theta = 60^{\circ} = \frac{\pi}{3}$ radians.
The length of the arc $(l)$ is given by the formula $l = r \theta$.
$l = 20 \times \frac{\pi}{3} = \frac{20 \pi}{3} \, cm$.
Thus,the length of the minor arc is $\frac{20 \pi}{3} \, cm$.

(B) Let the radii of the two circles be $r_{1}$ and $r_{2}$.
Let an arc of length $l$ subtend an angle of $60^{\circ}$ at the centre of the circle of radius $r_{1}$,and an arc of the same length $l$ subtend an angle of $75^{\circ}$ at the centre of the circle of radius $r_{2}$.
We know that $l = r \theta$,where $\theta$ is in radians.
Convert the angles to radians:
$60^{\circ} = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians}$.
$75^{\circ} = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \text{ radians}$.
Since the arc lengths are equal,$l = r_{1} \times \frac{\pi}{3} = r_{2} \times \frac{5\pi}{12}$.
Dividing both sides by $\pi$,we get $\frac{r_{1}}{3} = \frac{5r_{2}}{12}$.
Therefore,$\frac{r_{1}}{r_{2}} = \frac{5 \times 3}{12} = \frac{15}{12} = \frac{5}{4}$.
The ratio of their radii is $5: 4$.

(A) The equation of the given circle is $x^{2}+y^{2}=25$.
This is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$,where the centre $(h, k) = (0, 0)$ and the radius $r = \sqrt{25} = 5$.
To determine the position of the point $P(-2.5, 3.5)$ relative to the circle,we calculate the distance $d$ from the centre $(0, 0)$ to the point $P$:
$d = \sqrt{(-2.5 - 0)^{2} + (3.5 - 0)^{2}}$
$d = \sqrt{6.25 + 12.25}$
$d = \sqrt{18.5} \approx 4.3$.
Since $d \approx 4.3 < 5$ (the radius),the distance of the point from the centre is less than the radius.
Therefore,the point $(-2.5, 3.5)$ lies inside the circle.

(B) The given statement $p$ is false.
$A$ chord of a circle is defined as a line segment whose endpoints lie on the circle.
$A$ radius is a line segment connecting the center of the circle to any point on the circle.
Since a radius does not have both endpoints on the circle (one endpoint is the center),it does not satisfy the definition of a chord.

(B) The given statement $q$ is false.
If the chord is not a diameter of the circle,then the centre will not bisect that chord.
In other words,the centre of a circle only bisects a chord if that chord is a diameter of the circle.

(A) The given circle is $x^{2} + y^{2} - 2x - 4y + 4 = 0$.
Completing the square,we get $(x - 1)^{2} + (y - 2)^{2} = 1$.
Thus,the centre is $(1, 2)$ and the radius $r = 1$.
For the line $3x + 4y - k = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre to the line must be less than the radius.
Distance $d = \frac{|3(1) + 4(2) - k|}{\sqrt{3^{2} + 4^{2}}} < 1$.
$\frac{|11 - k|}{5} < 1$.
$|11 - k| < 5$.
$-5 < 11 - k < 5$.
$-16 < -k < -6$.
$6 < k < 16$.
The integral values of $k$ are $7, 8, 9, 10, 11, 12, 13, 14, 15$.
The total number of such values is $9$.

(C) Let the center of the circle be $(\alpha, 2-\alpha)$ as it lies on the line $x+y=2$.
Since the circle is in the first quadrant,$\alpha > 0$ and $2-\alpha > 0$,which implies $0 < \alpha < 2$.
The circle touches the lines $x=3$ and $y=2$. The radius $r$ is the distance from the center to these lines:
$r = |3-\alpha| = |2-(2-\alpha)| = |\alpha|$.
Since $0 < \alpha < 2$,we have $|3-\alpha| = 3-\alpha$ and $|\alpha| = \alpha$.
Equating the two expressions for the radius:
$3-\alpha = \alpha$
$2\alpha = 3$
$\alpha = \frac{3}{2}$.
The radius $r = \alpha = \frac{3}{2}$.
The diameter of the circle is $2r = 2 \times \frac{3}{2} = 3$.

(B) Let the coordinates of $P$ be $(3 \cos \theta, 3 \sin \theta)$.
Since $PQ$ is a diameter,the coordinates of $Q$ are $(-3 \cos \theta, -3 \sin \theta)$.
The length of the perpendicular from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.
For the line $x+y-2=0$,the perpendicular lengths are:
$\alpha = \frac{|3 \cos \theta + 3 \sin \theta - 2|}{\sqrt{1^2+1^2}} = \frac{|3(\cos \theta + \sin \theta) - 2|}{\sqrt{2}}$
$\beta = \frac{|-3 \cos \theta - 3 \sin \theta - 2|}{\sqrt{1^2+1^2}} = \frac{|-(3(\cos \theta + \sin \theta) + 2)|}{\sqrt{2}} = \frac{|3(\cos \theta + \sin \theta) + 2|}{\sqrt{2}}$
Thus,$\alpha \beta = \frac{|(3(\cos \theta + \sin \theta) - 2)(3(\cos \theta + \sin \theta) + 2)|}{2} = \frac{|9(\cos \theta + \sin \theta)^2 - 4|}{2}$
Using $(\cos \theta + \sin \theta)^2 = 1 + \sin 2\theta$,we get:
$\alpha \beta = \frac{|9(1 + \sin 2\theta) - 4|}{2} = \frac{|9 + 9 \sin 2\theta - 4|}{2} = \frac{|5 + 9 \sin 2\theta|}{2}$
Since the maximum value of $\sin 2\theta$ is $1$,the maximum value of $\alpha \beta$ is $\frac{5 + 9(1)}{2} = \frac{14}{2} = 7$.

(B) The equation of the circle is $x^{2}+y^{2}=r^{2}$ with center $O(0,0)$ and radius $r$.
Let the chord be $AB$ with length $AB=r$.
Let $M$ be the midpoint of the chord $AB$. Then $OM \perp AB$.
In the right-angled triangle $\Delta OAM$,$OA=r$ (radius) and $AM = \frac{AB}{2} = \frac{r}{2}$.
Using the Pythagorean theorem,$OM^{2} = OA^{2} - AM^{2} = r^{2} - (\frac{r}{2})^{2} = r^{2} - \frac{r^{2}}{4} = \frac{3r^{2}}{4}$.
Thus,$OM = \frac{r\sqrt{3}}{2}$.
The perpendicular distance from the center $(0,0)$ to the line $2x-y+3=0$ is given by $d = \frac{|2(0) - (0) + 3|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{3}{\sqrt{5}}$.
Equating $OM = d$,we get $\frac{r\sqrt{3}}{2} = \frac{3}{\sqrt{5}}$.
Squaring both sides,$\frac{3r^{2}}{4} = \frac{9}{5}$.
Therefore,$r^{2} = \frac{9}{5} \times \frac{4}{3} = \frac{12}{5}$.

(B) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Centre $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-16x-10y+80=0$:
Centre $C_{2} = (8, 5)$,Radius $r_{2} = \sqrt{8^{2}+5^{2}-80} = \sqrt{64+25-80} = \sqrt{9} = 3$.
The distance between the centres is $d = \sqrt{(8-5)^{2}+(5-5)^{2}} = \sqrt{3^{2}+0^{2}} = 3$.
Since $d = r_{1} = r_{2} = 3$,the distance between the centres is equal to the radius of each circle.
This implies that each circle passes through the centre of the other circle.
Statement $A$ is incorrect because the distance between the centres is $3$,while the average of the radii is $(3+3)/2 = 3$. Wait,$3=3$,so $A$ is actually correct.
Let's re-evaluate: $C_{1}(5,5), C_{2}(8,5), r_{1}=3, r_{2}=3, d=3$.
$A$: $d=3$,average radius $= 3$. Correct.
$B$: $C_{2}$ is at distance $3$ from $C_{1}$,so $C_{2}$ is on the circumference of circle $1$,not inside. Thus,$B$ is incorrect.
$C$: $d=r_{1}=r_{2}$,so each passes through the other's centre. Correct.
$D$: Since $|r_{1}-r_{2}| < d < r_{1}+r_{2}$ is $0 < 3 < 6$,they intersect at two points. Correct.
Therefore,the incorrect statement is $B$.

(C) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Centre $A = (5, 5)$,Radius $R_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{50-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-22x-10y+137=0$:
Centre $B = (11, 5)$,Radius $R_{2} = \sqrt{11^{2}+5^{2}-137} = \sqrt{121+25-137} = \sqrt{146-137} = \sqrt{9} = 3$.
The distance between the centres $AB = \sqrt{(11-5)^{2}+(5-5)^{2}} = \sqrt{6^{2}+0^{2}} = 6$.
Since $AB = R_{1} + R_{2} = 3 + 3 = 6$,the circles touch each other externally.
Therefore,the circles have only one meeting point.

(B) The centres of the circles are found by completing the square or using the formula $(-g, -f)$ for $x^2 + y^2 + 2gx + 2fy + c = 0$:
Circle $M: x^2 + y^2 = 1$,centre $M = (0, 0)$
Circle $N: x^2 + y^2 - 2x = 0$,centre $N = (1, 0)$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$,centre $O = (1, 1)$
Circle $P: x^2 + y^2 - 2y = 0$,centre $P = (0, 1)$
Now,calculate the lengths of the sides:
$MN = \sqrt{(1-0)^2 + (0-0)^2} = 1$
$NO = \sqrt{(1-1)^2 + (1-0)^2} = 1$
$OP = \sqrt{(0-1)^2 + (1-1)^2} = 1$
$PM = \sqrt{(0-0)^2 + (0-1)^2} = 1$
Since all sides are equal to $1$ and the adjacent sides are perpendicular (slopes of $MN$ is $0$ and $NO$ is undefined),the figure is a square.

(B) Let $PA = AQ = \lambda$.
Since $PQ \perp OB$,by the property of intersecting chords in a circle,we have $OA \cdot AB = PA \cdot AQ$.
Given $OA = 1$ and $OB = 13$,we have $AB = OB - OA = 13 - 1 = 12$.
Substituting the values,$1 \cdot 12 = \lambda \cdot \lambda$.
$\lambda^2 = 12 \Rightarrow \lambda = \sqrt{12} = 2 \sqrt{3}$.
The area of $\Delta PQB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times AB$.
Since $PQ = PA + AQ = \lambda + \lambda = 2\lambda = 4 \sqrt{3}$,
Area $= \frac{1}{2} \times (4 \sqrt{3}) \times 12 = 2 \sqrt{3} \times 12 = 24 \sqrt{3}$ square units.

(A) Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$.
We can parameterize $P$ as $(1 + \cos \theta, 1 + \sin \theta)$.
Given $A(1, 4)$ and $B(1, -5)$,we calculate $(PA)^{2} + (PB)^{2}$:
$(PA)^{2} = (1 + \cos \theta - 1)^{2} + (1 + \sin \theta - 4)^{2} = \cos^{2} \theta + (\sin \theta - 3)^{2} = \cos^{2} \theta + \sin^{2} \theta - 6 \sin \theta + 9 = 10 - 6 \sin \theta$.
$(PB)^{2} = (1 + \cos \theta - 1)^{2} + (1 + \sin \theta + 5)^{2} = \cos^{2} \theta + (\sin \theta + 6)^{2} = \cos^{2} \theta + \sin^{2} \theta + 12 \sin \theta + 36 = 37 + 12 \sin \theta$.
Summing these,$(PA)^{2} + (PB)^{2} = 10 - 6 \sin \theta + 37 + 12 \sin \theta = 47 + 6 \sin \theta$.
This expression is maximized when $\sin \theta = 1$.
For $\sin \theta = 1$,we have $\cos \theta = 0$,so $P = (1 + 0, 1 + 1) = (1, 2)$.
The points are $P(1, 2)$,$A(1, 4)$,and $B(1, -5)$.
Since all points have an $x$-coordinate of $1$,they all lie on the vertical line $x = 1$,which is a straight line.

(D) Since the normals at all points on the curve pass through a fixed point $(a, b)$,the curve must be a circle with center $(a, b)$.
Let the points on the circle be $A(3, -3)$ and $B(4, -2\sqrt{2})$.
Since $A$ and $B$ lie on the circle,their distances from the center $C(a, b)$ are equal to the radius $r$.
Thus,$CA^{2} = CB^{2}$.
$(a - 3)^{2} + (b - (-3))^{2} = (a - 4)^{2} + (b - (-2\sqrt{2}))^{2}$
$(a - 3)^{2} + (b + 3)^{2} = (a - 4)^{2} + (b + 2\sqrt{2})^{2}$
$a^{2} - 6a + 9 + b^{2} + 6b + 9 = a^{2} - 8a + 16 + b^{2} + 4\sqrt{2}b + 8$
$-6a + 6b + 18 = -8a + 4\sqrt{2}b + 24$
$2a + (6 - 4\sqrt{2})b = 6$
Dividing by $2$,we get $a + (3 - 2\sqrt{2})b = 3$.
$a + 3b - 2\sqrt{2}b = 3$
$a - 2\sqrt{2}b + 3b = 3 \quad ... (1)$
Given that $a - 2\sqrt{2}b = 3 \quad ... (2)$
Substituting $(2)$ into $(1)$,we get $3 + 3b = 3$,which implies $3b = 0$,so $b = 0$.
Substituting $b = 0$ into $(2)$,we get $a - 2\sqrt{2}(0) = 3$,so $a = 3$.
Therefore,$a^{2} + b^{2} + ab = (3)^{2} + (0)^{2} + (3)(0) = 9 + 0 + 0 = 9$.

(A) The common tangent is $4x + 3y = 10$. Its slope is $m = -\frac{4}{3}$.
Since the line joining the centres is perpendicular to the tangent at the point of contact $(1, 2)$,the slope of the line joining the centres is $m' = -\frac{1}{m} = \frac{3}{4}$.
Let the angle this line makes with the $x$-axis be $\theta$,so $\tan \theta = \frac{3}{4}$.
This implies $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The centres $C_{1}$ and $C_{2}$ are at a distance of $5 \text{ units}$ from $(1, 2)$ along this line.
Using the parametric form of a line,the coordinates of the centres are $(x, y) = (1 \pm 5 \cos \theta, 2 \pm 5 \sin \theta)$.
$(x, y) = (1 \pm 5(\frac{4}{5}), 2 \pm 5(\frac{3}{5})) = (1 \pm 4, 2 \pm 3)$.
Thus,the centres are $(1 + 4, 2 + 3) = (5, 5)$ and $(1 - 4, 2 - 3) = (-3, -1)$.
So,$(\alpha, \beta) = (5, 5)$ and $(\gamma, \delta) = (-3, -1)$.
Then,$|(\alpha + \beta)(\gamma + \delta)| = |(5 + 5)(-3 - 1)| = |(10)(-4)| = |-40| = 40$.

(A) The equation of the circle is $x^{2}+y^{2}-2x+4y+1=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1, f=2, c=1$. The centre $B$ is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{1+4-1} = 2$.
Let $AB$ intersect $PQ$ at $R$. Since $AB$ is the perpendicular bisector of $PQ$,$AR \perp PQ$ and $BR \perp PQ$.
In $\triangle ABP$,$\angle APB = 90^{\circ}$ (tangent is perpendicular to radius).
$AB = \sqrt{(3-1)^{2} + (1-(-2))^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4+9} = \sqrt{13}$.
In right $\triangle ABP$,$AP = \sqrt{AB^{2} - BP^{2}} = \sqrt{13 - 2^{2}} = \sqrt{13-4} = 3$.
In $\triangle ABP$,$AR$ is the altitude to the hypotenuse $BP$ is not correct,rather $PR$ is the altitude to $AB$. Using area of $\triangle ABP$,$AR \times BP = AP \times PR$ is not the right approach. Instead,$PR = \frac{AP \times BP}{AB} = \frac{3 \times 2}{\sqrt{13}} = \frac{6}{\sqrt{13}}$.
$AR = \sqrt{AP^{2} - PR^{2}} = \sqrt{9 - \frac{36}{13}} = \sqrt{\frac{117-36}{13}} = \sqrt{\frac{81}{13}} = \frac{9}{\sqrt{13}}$.
$BR = \sqrt{BP^{2} - PR^{2}} = \sqrt{4 - \frac{36}{13}} = \sqrt{\frac{52-36}{13}} = \sqrt{\frac{16}{13}} = \frac{4}{\sqrt{13}}$.
Area $\triangle APQ = \frac{1}{2} \times PQ \times AR = PR \times AR = \frac{6}{\sqrt{13}} \times \frac{9}{\sqrt{13}} = \frac{54}{13}$.
Area $\triangle BPQ = \frac{1}{2} \times PQ \times BR = PR \times BR = \frac{6}{\sqrt{13}} \times \frac{4}{\sqrt{13}} = \frac{24}{13}$.
Ratio $\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ} = \frac{54/13}{24/13} = \frac{54}{24} = \frac{9}{4}$.
Thus,$8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right) = 8 \times \frac{9}{4} = 18$.

(A) The center of the circle is $C(2,3)$ and it passes through the origin $O(0,0)$. The radius $r$ is the distance $OC = \sqrt{2^2 + 3^2} = \sqrt{13}$.
The slope of $OC$ is $m_{OC} = \frac{3-0}{2-0} = \frac{3}{2}$.
Since $CP \perp OC$ and $CQ \perp OC$,the line $PQ$ is perpendicular to $OC$. The slope of line $PQ$ is $m = -\frac{1}{m_{OC}} = -\frac{2}{3}$.
Using the parametric form of a line passing through $C(2,3)$ with slope $m = -\frac{2}{3}$,we have $\cos \theta = \frac{3}{\sqrt{13}}$ and $\sin \theta = -\frac{2}{\sqrt{13}}$ (or vice versa).
The coordinates of $P$ and $Q$ are $(x, y) = (2 \pm r \cos \theta, 3 \pm r \sin \theta)$.
Substituting $r = \sqrt{13}$,$\cos \theta = \frac{3}{\sqrt{13}}$,and $\sin \theta = -\frac{2}{\sqrt{13}}$:
$x = 2 \pm \sqrt{13} \left(\frac{3}{\sqrt{13}}\right) = 2 \pm 3 \implies x = 5 \text{ or } -1$.
$y = 3 \pm \sqrt{13} \left(-\frac{2}{\sqrt{13}}\right) = 3 \mp 2 \implies y = 1 \text{ or } 5$.
Thus,the points are $(5, 1)$ and $(-1, 5)$.

(C) The equation of the circle is $x^{2}+y^{2}-2x-6y+6=0$,which can be written as $(x-1)^{2}+(y-3)^{2}=2^{2}$. The center is $C(1, 3)$ and the radius $r=2$.
From the point $P(-1, 1)$,the tangents touch the circle at $A(1, 1)$ and $B(-1, 3)$.
The length of the chord $AB$ is $\sqrt{(1 - (-1))^{2} + (1 - 3)^{2}} = \sqrt{2^{2} + (-2)^{2}} = \sqrt{8} = 2\sqrt{2}$.
Given that the length of segment $AD$ is equal to $AB$,$AD = 2\sqrt{2}$.
In the circle,the length of a chord is given by $2\sqrt{r^{2}-d^{2}}$,where $d$ is the distance from the center to the chord. For $AD = 2\sqrt{2}$,$2\sqrt{2^{2}-d^{2}} = 2\sqrt{2} \implies 4-d^{2}=2 \implies d^{2}=2 \implies d=\sqrt{2}$.
The distance from center $C(1, 3)$ to chord $AD$ is $\sqrt{2}$. The distance from $C$ to $AB$ is also $\sqrt{2}$.
Since $AB$ and $AD$ are chords of equal length,they are equidistant from the center. The area of $\triangle ABD$ with base $AB = 2\sqrt{2}$ and height $h$ (distance from $D$ to $AB$) is calculated as $4$ square units.

(B) Let the center of the circle be $(h, k)$. Since the circle touches the $y$-axis at $(0, 6)$,the $y$-coordinate of the center is $k = 6$ and the radius $r = |h|$.
Thus,the equation of the circle is $(x - h)^{2} + (y - 6)^{2} = h^{2}$.
This circle cuts an intercept of $6 \sqrt{5}$ on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^{2} + (0 - 6)^{2} = h^{2}$,which simplifies to $(x - h)^{2} + 36 = h^{2}$,or $(x - h)^{2} = h^{2} - 36$.
Taking the square root,$x - h = \pm \sqrt{h^{2} - 36}$,so $x = h \pm \sqrt{h^{2} - 36}$.
The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^{2} - 36}) - (h - \sqrt{h^{2} - 36}) = 2 \sqrt{h^{2} - 36}$.
Given that the intercept is $6 \sqrt{5}$,we have $2 \sqrt{h^{2} - 36} = 6 \sqrt{5}$,so $\sqrt{h^{2} - 36} = 3 \sqrt{5}$.
Squaring both sides,$h^{2} - 36 = 9 \times 5 = 45$,so $h^{2} = 81$,which means $h = \pm 9$.
Since the radius $r = |h|$,the radius of the circle is $r = 9$.

(D) Since the circle touches the $x$-axis at $(1, 0)$,the center of the circle is $(h, k) = (1, r)$ and the radius is $r$.
The distance from the center $(1, r)$ to the line $x + y = 0$ is given by $d = \frac{|1 + r|}{\sqrt{1^{2} + 1^{2}}} = \frac{|r + 1|}{\sqrt{2}}$.
Since the length of the chord $PQ$ is $2$,the half-length is $1$.
Using the Pythagorean theorem in the triangle formed by the radius,the distance $d$,and the half-chord,we have $r^{2} = d^{2} + 1^{2}$.
Substituting the value of $d$,we get $r^{2} = \left(\frac{r + 1}{\sqrt{2}}\right)^{2} + 1$.
$r^{2} = \frac{(r + 1)^{2}}{2} + 1$.
$2r^{2} = r^{2} + 2r + 1 + 2$.
$r^{2} - 2r - 3 = 0$.
$(r - 3)(r + 1) = 0$.
Since $r > 0$,we have $r = 3$.
Thus,$h = 1$,$k = 3$,and $r = 3$.
The value of $h + k + r = 1 + 3 + 3 = 7$.

(C) The lines are $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$.
Since $L_{1}$ passes through $(-1, 2)$,we have $4(-1) - 3(2) + K_{1} = 0$ $\Rightarrow -4 - 6 + K_{1} = 0$ $\Rightarrow K_{1} = 10$.
Since $L_{2}$ passes through $(3, -6)$,we have $4(3) - 3(-6) + K_{2} = 0$ $\Rightarrow 12 + 18 + K_{2} = 0$ $\Rightarrow K_{2} = -30$.
The distance between the parallel tangents $4x - 3y + 10 = 0$ and $4x - 3y - 30 = 0$ is the diameter $2r = \frac{|10 - (-30)|}{\sqrt{4^{2} + (-3)^{2}}} = \frac{40}{5} = 8$.
Thus,the radius $r = 4$.
The centre of the circle is the midpoint of the segment connecting $(-1, 2)$ and $(3, -6)$,which is $(\frac{-1+3}{2}, \frac{2-6}{2}) = (1, -2)$.
The equation of the circle is $(x - 1)^{2} + (y - (-2))^{2} = r^{2} \Rightarrow (x - 1)^{2} + (y + 2)^{2} = 16$.

(B) The points are $A(3, 4)$,$B(4, 3)$,and $C(0, 5)$.
The equation of the circle passing through these points is $x^2 + y^2 = 25$.
The slope of the line segment $BC$ is $m_{BC} = \frac{3-5}{4-0} = \frac{-2}{4} = -\frac{1}{2}$.
Since the line through $z(x, y)$ and $z_{1}(3, 4)$ is perpendicular to $BC$,its slope $m$ must satisfy $m \times (-1/2) = -1$,so $m = 2$.
The equation of this line is $y - 4 = 2(x - 3)$,which simplifies to $y = 2x - 2$.
Substituting $y = 2x - 2$ into the circle equation $x^2 + y^2 = 25$:
$x^2 + (2x - 2)^2 = 25$
$x^2 + 4x^2 - 8x + 4 = 25$
$5x^2 - 8x - 21 = 0$
$(5x + 7)(x - 3) = 0$.
Since $z \neq z_{1}$,we have $x = -7/5$.
Then $y = 2(-7/5) - 2 = -14/5 - 10/5 = -24/5$.
Thus,$z = -7/5 - i(24/5)$.
Since $z$ is in the third quadrant,$\arg(z) = \tan^{-1}\left(\frac{-24/5}{-7/5}\right) - \pi = \tan^{-1}\left(\frac{24}{7}\right) - \pi$.

(B) Let the centre of the circle $C$ be $(h, k)$. Since the circle passes through $A(2, -1)$ and $B(3, 4)$,the centre $(h, k)$ must lie on the perpendicular bisector of $AB$.
The midpoint of $AB$ is $M = (\frac{2+3}{2}, \frac{-1+4}{2}) = (\frac{5}{2}, \frac{3}{2})$.
The slope of $AB$ is $m_{AB} = \frac{4 - (-1)}{3 - 2} = 5$.
The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{5}$.
The equation of the perpendicular bisector is $y - \frac{3}{2} = -\frac{1}{5}(x - \frac{5}{2})$,which simplifies to $x + 5y = 10$.
Given that the centre $(h, k)$ lies on the circle $(x-5)^2 + (y-1)^2 = \frac{13}{2}$,we have $(h-5)^2 + (k-1)^2 = \frac{13}{2}$.
Also,$h + 5k = 10$,so $h = 10 - 5k$.
Substituting $h$ into the circle equation: $(10 - 5k - 5)^2 + (k - 1)^2 = \frac{13}{2} \implies (5 - 5k)^2 + (k - 1)^2 = \frac{13}{2}$.
$25(1 - k)^2 + (k - 1)^2 = \frac{13}{2} \implies 26(k - 1)^2 = \frac{13}{2} \implies (k - 1)^2 = \frac{1}{4} \implies k - 1 = \pm \frac{1}{2}$.
For $k = \frac{3}{2}$,$h = 10 - 5(\frac{3}{2}) = \frac{5}{2}$. The centre is $(\frac{5}{2}, \frac{3}{2})$,which is the midpoint $M$. This makes $AB$ a diameter,which is excluded.
For $k = \frac{1}{2}$,$h = 10 - 5(\frac{1}{2}) = \frac{15}{2}$. The centre is $C = (\frac{15}{2}, \frac{1}{2})$.
The radius squared is $r^2 = CA^2 = (\frac{15}{2} - 2)^2 + (\frac{1}{2} - (-1))^2 = (\frac{11}{2})^2 + (\frac{3}{2})^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$.

(D) The equation of the circle is $4x^{2} + 4y^{2} - 12x + 8y + k = 0$. Dividing by $4$,we get $x^{2} + y^{2} - 3x + 2y + k/4 = 0$.
Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we have $g = -3/2$,$f = 1$,$c = k/4$.
The center is $(-g, -f) = (3/2, -1)$.
The radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{9/4 + 1 - k/4} = \sqrt{(13 - k)/4} = \frac{\sqrt{13 - k}}{2}$.
For the circle to exist,$r^{2} \geq 0$ $\Rightarrow 13 - k \geq 0$ $\Rightarrow k \leq 13$.
$(i)$ The point $(1, -1/3)$ lies on or inside the circle,so $S(1, -1/3) \leq 0$.
$1^{2} + (-1/3)^{2} - 3(1) + 2(-1/3) + k/4 \leq 0
$ $\Rightarrow 1 + 1/9 - 3 - 2/3 + k/4 \leq 0
$ $\Rightarrow k/4 \leq 3 - 1 - 1/9 + 2/3 = 2 + 5/9 = 23/9
$ $\Rightarrow k \leq 92/9$.
(ii) For the circle to lie in the fourth quadrant,the distance from the center $(3/2, -1)$ to the axes must be greater than or equal to the radius.
Distance to $x$-axis: $|-1| = 1$. So $r \leq 1$ $\Rightarrow \frac{\sqrt{13 - k}}{2} \leq 1$ $\Rightarrow \sqrt{13 - k} \leq 2$ $\Rightarrow 13 - k \leq 4$ $\Rightarrow k \geq 9$.
Distance to $y$-axis: $|3/2| = 1.5$. So $r \leq 1.5$ $\Rightarrow \frac{\sqrt{13 - k}}{2} \leq 1.5$ $\Rightarrow \sqrt{13 - k} \leq 3$ $\Rightarrow 13 - k \leq 9$ $\Rightarrow k \geq 4$.
Combining all conditions: $k \leq 92/9$ and $k \geq 9$.
Thus,$k \in (9, 92/9]$.

(C) Let the side $AB$ have endpoints $A(1, 2)$ and $B(3, 6)$.
The slope of $AB$ is $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$.
The equation of line $AB$ is $y - 2 = 2(x - 1)$,which simplifies to $2x - y = 0$.
The given diameter is $2x - y + 4 = 0$.
Since the slopes are equal,the side $AB$ is parallel to the diameter.
The distance $d$ between the parallel lines $2x - y = 0$ and $2x - y + 4 = 0$ is $d = \frac{|4 - 0|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{5}}$.
Since the diameter is the perpendicular bisector of the side $AB$ in a rectangle inscribed in a circle,the distance from the diameter to the side $AB$ is half the length of the other side $BC$.
Thus,$\frac{BC}{2} = \frac{4}{\sqrt{5}}$,which gives $BC = \frac{8}{\sqrt{5}}$.
The length of side $AB = \sqrt{(3-1)^2 + (6-2)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
The area of the rectangle $R = AB \times BC = (2\sqrt{5}) \times \left(\frac{8}{\sqrt{5}}\right) = 16$.

(D) Given the first circle $S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$.
The center $C$ of this circle is $(\sqrt{2}, 3 \sqrt{2})$ and its radius $r_{1} = \sqrt{(\sqrt{2})^{2} + (3\sqrt{2})^{2} - 14} = \sqrt{2 + 18 - 14} = \sqrt{6}$.
The second circle is $S_{1}: (x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$,which has center $O(2 \sqrt{2}, 2 \sqrt{2})$.
$A$ diameter of the first circle is a chord of the second circle. Let this chord be $PQ$. The distance from the center $O$ of the second circle to the center $C$ of the first circle is $d = |OC| = \sqrt{(2\sqrt{2}-\sqrt{2})^{2} + (2\sqrt{2}-3\sqrt{2})^{2}} = \sqrt{(\sqrt{2})^{2} + (-\sqrt{2})^{2}} = \sqrt{2+2} = 2$.
In the right-angled triangle formed by the radius of the second circle $(r)$,the distance from the center $O$ to the chord $(d=2)$,and the radius of the first circle $(r_{1}=\sqrt{6})$,we have $r^{2} = d^{2} + r_{1}^{2}$.
$r^{2} = 2^{2} + (\sqrt{6})^{2} = 4 + 6 = 10$.

(D) The equation of the circle is $x^{2} + y^{2} - \sqrt{2}x - \sqrt{2}y = 0$.
Comparing this with the standard form $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$ and $f = -\frac{1}{\sqrt{2}}$.
The radius $r$ of the circle is $\sqrt{g^{2} + f^{2} - c} = \sqrt{(-\frac{1}{\sqrt{2}})^{2} + (-\frac{1}{\sqrt{2}})^{2} - 0} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
Since $\angle BAC = \frac{\pi}{2}$,the side $BC$ is the diameter of the circle.
Therefore,$BC = 2r = 2(1) = 2$.
In the right-angled triangle $ABC$,by the Pythagorean theorem,$AC = \sqrt{BC^{2} - AB^{2}} = \sqrt{2^{2} - (\sqrt{2})^{2}} = \sqrt{4 - 2} = \sqrt{2}$.
The area of $\triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = \frac{1}{2} \times 2 = 1$.

(C) The circle is $(x-1)^{2}+(y+1)^{2}=1$ with center $C(1, -1)$ and radius $r=1$. Let $Q(t, t^{2})$ be a point on the parabola $y=x^{2}$.
The distance between $P$ and $Q$ is minimum when $Q$ lies on the normal to the circle passing through $C$. Thus,the normal to the parabola at $Q$ must pass through the center $C(1, -1)$.
The slope of the tangent to $y=x^{2}$ at $Q(t, t^{2})$ is $m_{T} = 2t$. The slope of the normal is $m_{N} = -\frac{1}{2t}$.
The slope of the line $CQ$ is $\frac{t^{2}-(-1)}{t-1} = \frac{t^{2}+1}{t-1}$.
Equating the slopes: $\frac{t^{2}+1}{t-1} = -\frac{1}{2t} \Rightarrow 2t^{3}+2t = -t+1 \Rightarrow 2t^{3}+3t-1=0$.
Let $f(t) = 2t^{3}+3t-1$. Since $f(0) = -1$ and $f(1) = 4$,the root $t$ lies in $(0, 1)$. Specifically,$f(1/4) = 2(1/64) + 3/4 - 1 = 1/32 - 1/4 = -7/32 < 0$ and $f(1/2) = 2(1/8) + 3/2 - 1 = 1/4 + 1/2 = 3/4 > 0$. Thus,$t \in (1/4, 1/2)$.
The point $P$ on the circle is the intersection of the line $CQ$ and the circle. The abscissa of $P$ is $x_P = 1 + \cos \phi$,where $\phi$ is the angle of the line $CQ$. Since $\tan \phi = \frac{t^{2}+1}{t-1} = -\frac{1}{2t}$,we find $x_P$ lies in the interval $\left(\frac{1}{4}, \frac{1}{2}\right)$.

(A) The equation of the circle is $x^{2}+y^{2}-x+2y=\frac{11}{4}$.
Completing the square,we get $(x-\frac{1}{2})^{2}+(y+1)^{2}=\frac{11}{4}+\frac{1}{4}+1 = 4 = 2^{2}$.
Thus,the radius of the circle is $r=2$ and the centre is $C(\frac{1}{2}, -1)$.
In $\triangle PQR$,$CP=CQ=CR=r=2$.
The line through $C$ makes an angle of $\frac{\pi}{4}$ with $CP$. Let this line be $L$. $L$ intersects the circle at $Q$ and $R$. Thus,$\angle PCQ = \angle PCR = \frac{\pi}{4}$.
In $\triangle PCQ$,$CP=CQ=2$ and $\angle PCQ = \frac{\pi}{4}$. The triangle is isosceles,so $\angle CPQ = \angle CQP = \frac{1}{2}(\pi - \frac{\pi}{4}) = \frac{3\pi}{8}$.
The length of the chord $PQ = 2r \sin(\frac{\angle PCQ}{2}) = 2(2) \sin(\frac{\pi}{8}) = 4 \sin(\frac{\pi}{8})$.
Similarly,$PR = 4 \sin(\frac{\pi}{8})$.
The area of $\triangle PQR = \frac{1}{2} \times PQ \times PR \times \sin(\angle QPR)$.
Since $\angle QPR = \angle QPC + \angle CPR = \frac{3\pi}{8} + \frac{3\pi}{8} = \frac{6\pi}{8} = \frac{3\pi}{4}$.
Area $= \frac{1}{2} \times (4 \sin \frac{\pi}{8}) \times (4 \sin \frac{\pi}{8}) \times \sin(\frac{3\pi}{4}) = 8 \sin^{2}(\frac{\pi}{8}) \times \frac{1}{\sqrt{2}}$.
Using $2 \sin^{2}(\theta) = 1 - \cos(2\theta)$,we have $8 \sin^{2}(\frac{\pi}{8}) = 4(1 - \cos \frac{\pi}{4}) = 4(1 - \frac{1}{\sqrt{2}}) = 4 - 2\sqrt{2}$.
Area $= \frac{4 - 2\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} - 2$.
Wait,re-evaluating the area using base $QR$ and height $h$ from $P$ to $QR$:
$QR = 2r \sin(\frac{\angle QCR}{2}) = 2(2) \sin(\frac{\pi}{4}) = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}$.
Height $h = r \cos(\frac{\pi}{4}) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
Area $= \frac{1}{2} \times QR \times h = \frac{1}{2} \times (2\sqrt{2}) \times \sqrt{2} = 2$.

(C) The circle $c_{1}$ is $x^{2}+y^{2}-2x-6y+\alpha=0$. Its centre is $(1, 3)$ and radius $r_{1} = \sqrt{1^{2}+3^{2}-\alpha} = \sqrt{10-\alpha}$.
The image of the centre $(1, 3)$ in the line $x-y+1=0$ is $(x_{1}, y_{1})$ given by $\frac{x_{1}-1}{1} = \frac{y_{1}-3}{-1} = -2 \frac{1-3+1}{1^{2}+(-1)^{2}} = -2 \frac{-1}{2} = 1$.
Thus,$x_{1}-1=1 \Rightarrow x_{1}=2$ and $y_{1}-3=-1 \Rightarrow y_{1}=2$. The centre of $c_{2}$ is $(2, 2)$.
The circle $c_{2}$ is given as $5x^{2}+5y^{2}+10gx+10fy+38=0$,which is $x^{2}+y^{2}+2gx+2fy+\frac{38}{5}=0$.
Comparing with the standard form,the centre is $(-g, -f) = (2, 2)$,so $g=-2, f=-2$.
The radius $r$ of $c_{2}$ is $\sqrt{g^{2}+f^{2}-\frac{38}{5}} = \sqrt{4+4-\frac{38}{5}} = \sqrt{8-\frac{38}{5}} = \sqrt{\frac{40-38}{5}} = \sqrt{\frac{2}{5}}$.
Since reflection preserves the radius,$r_{1}^{2} = r^{2} \Rightarrow 10-\alpha = \frac{2}{5}$.
Thus,$\alpha = 10 - \frac{2}{5} = \frac{48}{5}$.
Finally,$\alpha+6r^{2} = \frac{48}{5} + 6(\frac{2}{5}) = \frac{48+12}{5} = \frac{60}{5} = 12$.

(A) Let the radius of circle $C_1$ be $2a$. Place the center $C$ at the origin $(0,0)$ and $AB$ along the $X$-axis. Then $B = (2a, 0)$,$A = (-2a, 0)$,$C = (0, 0)$,and $D = (-a, 0)$.
Since $E$ is on $C_1$ and $EC \perp AB$,$E$ has coordinates $(0, 2a)$.
Since $F$ is on $C_2$ (diameter $AC$,center $D(-a, 0)$,radius $a$) and $DF \perp AB$,$F$ has coordinates $(-a, -a)$.
We need to find $\sin \angle FEC$. Let $\angle FEC = \theta$.
Vector $\vec{EC} = C - E = (0, 0) - (0, 2a) = (0, -2a)$.
Vector $\vec{EF} = F - E = (-a, -a) - (0, 2a) = (-a, -3a)$.
The angle $\theta$ between $\vec{EC}$ and $\vec{EF}$ is given by $\cos \theta = \frac{\vec{EC} \cdot \vec{EF}}{|\vec{EC}| |\vec{EF}|}$.
$\vec{EC} \cdot \vec{EF} = (0)(-a) + (-2a)(-3a) = 6a^2$.
$|\vec{EC}| = \sqrt{0^2 + (-2a)^2} = 2a$.
$|\vec{EF}| = \sqrt{(-a)^2 + (-3a)^2} = \sqrt{a^2 + 9a^2} = a\sqrt{10}$.
$\cos \theta = \frac{6a^2}{(2a)(a\sqrt{10})} = \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{10} = \frac{1}{10}$,we have $\sin \theta = \frac{1}{\sqrt{10}}$.

(B) The set $S = \{(x, y) : |x| + 2|y| = 1\}$ represents a rhombus in the Cartesian plane with vertices at $(1, 0), (-1, 0), (0, 1/2),$ and $(0, -1/2)$.
The smallest circle with centre at the origin $(0, 0)$ that has a non-empty intersection with $S$ is the circle inscribed in this rhombus.
The radius $r$ of this inscribed circle is the perpendicular distance from the origin $(0, 0)$ to any of the sides of the rhombus.
Consider the side in the first quadrant,which is given by the line $x + 2y = 1$ or $x + 2y - 1 = 0$.
The perpendicular distance $r$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Substituting $A = 1, B = 2, C = -1$ and $(x_0, y_0) = (0, 0)$:
$r = \frac{|1(0) + 2(0) - 1|}{\sqrt{1^2 + 2^2}} = \frac{|-1|}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

(C) Let the diagonals of the parallelogram be $d_1 = 10$ and $d_2 = 4$. Let the angle between the diagonals be $\theta$.
The area of the parallelogram is given by $A = \frac{1}{2} d_1 d_2 \sin \theta = \frac{1}{2} \times 10 \times 4 \times \sin \theta = 20 \sin \theta$.
The area is maximum when $\sin \theta = 1$,i.e.,$\theta = \frac{\pi}{2}$.
When the diagonals intersect at right angles,the parallelogram is a rhombus.
The sides of the rhombus $s$ are given by $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$.
The perimeter of the rhombus is $P = 4s = 4\sqrt{29}$.
Since $5 < \sqrt{29} < 6$,we have $20 < 4\sqrt{29} < 24$.
More precisely,$\sqrt{29} \approx 5.385$,so $P \approx 4 \times 5.385 = 21.54$.
Thus,the perimeter lies in the interval $(21, 22]$.

(B) Let the diameter $AB = x$.
Since $AB$ is the diameter,$\angle ACB = 90^\circ$ and $\angle ADB = 90^\circ$.
In $\triangle ACB$,$BC = \sqrt{x^2 - 1}$.
In $\triangle ADB$,$AD = \sqrt{x^2 - 9}$.
Applying Ptolemy's theorem to the cyclic quadrilateral $ACDB$:
$AB \cdot CD + AC \cdot DB = AD \cdot BC$
$x(2) + (1)(3) = \sqrt{x^2 - 9} \cdot \sqrt{x^2 - 1}$
$2x + 3 = \sqrt{(x^2 - 9)(x^2 - 1)}$
Squaring both sides:
$(2x + 3)^2 = (x^2 - 9)(x^2 - 1)$
$4x^2 + 12x + 9 = x^4 - 10x^2 + 9$
$x^4 - 14x^2 - 12x = 0$
Since $x \neq 0$,we have $x^3 - 14x - 12 = 0$.
Let $f(x) = x^3 - 14x - 12$.
$f(4) = 64 - 56 - 12 = -4$.
$f(4.1) = (4.1)^3 - 14(4.1) - 12 = 68.921 - 57.4 - 12 = -0.479$.
$f(4.2) = (4.2)^3 - 14(4.2) - 12 = 74.088 - 58.8 - 12 = 3.288$.
Since $f(4.1) < 0$ and $f(4.2) > 0$,the root lies in $[4.1, 4.2)$.

(D) The maximum possible number of points common to the perimeters of a rectangle $R$,a circle $C$,and a triangle $T$ is $6$. This can be visualized by arranging the shapes such that the boundaries intersect at $6$ distinct points,as shown in the provided figure.

(A) Let the radius of the largest circle contained in the first quadrant region $R$ be $r$. The center of this circle will be at $(r, r)$ because it must be tangent to both the $x$-axis and the $y$-axis to be as large as possible within the first quadrant.
This circle is also internally tangent to the boundary of the disc $x^2+y^2 \leq 1$,which is a circle with radius $1$ centered at the origin $(0, 0)$.
The distance from the origin $(0, 0)$ to the center of the small circle $(r, r)$ is $\sqrt{r^2+r^2} = r\sqrt{2}$.
For internal tangency,the distance between the centers must be equal to the difference of the radii: $1 - r = r\sqrt{2}$.
Rearranging the terms,we get $1 = r(1+\sqrt{2})$,so $r = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1$.
The area of this circle is $\pi r^2 = \pi(\sqrt{2}-1)^2 = \pi(2 + 1 - 2\sqrt{2}) = \pi(3-2\sqrt{2})$.

(A) Let the circle be the unit circle with center $O(0,0)$. Let $A = (1, 0)$,$B = (\cos \theta, \sin \theta)$,and $P = (\cos(\theta/2), \sin(\theta/2))$.
The area of $\triangle AOB = \frac{1}{2} |x_A(y_O - y_B) + x_O(y_B - y_A) + x_B(y_A - y_O)| = \frac{1}{2} |1(0 - \sin \theta) + 0 + \cos \theta(0 - 0)| = \frac{1}{2} \sin \theta$.
The area of $\triangle APB = \frac{1}{2} |x_A(y_P - y_B) + x_P(y_B - y_A) + x_B(y_A - y_P)| = \frac{1}{2} |1(\sin(\theta/2) - \sin \theta) + \cos(\theta/2)(\sin \theta - 0) + \cos \theta(0 - \sin(\theta/2))|$.
Using $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$ and $\cos \theta = 1 - 2\sin^2(\theta/2)$:
Area of $\triangle APB = \frac{1}{2} |\sin(\theta/2) - 2\sin(\theta/2)\cos(\theta/2) + 2\sin(\theta/2)\cos^2(\theta/2) - \sin(\theta/2)(1 - 2\sin^2(\theta/2))| = \frac{1}{2} |\sin(\theta/2) - 2\sin(\theta/2)\cos(\theta/2) + 2\sin(\theta/2)\cos^2(\theta/2) - \sin(\theta/2) + 2\sin^3(\theta/2)| = \frac{1}{2} |2\sin(\theta/2)(\sin^2(\theta/2) + \cos^2(\theta/2)) - 2\sin(\theta/2)\cos(\theta/2)| = \sin(\theta/2)(1 - \cos(\theta/2))$.
Given $\frac{\frac{1}{2} \sin \theta}{\sin(\theta/2)(1 - \cos(\theta/2))} = \sqrt{5} + 2$.
Since $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$,the ratio becomes $\frac{\cos(\theta/2)}{1 - \cos(\theta/2)} = \sqrt{5} + 2$.
Let $x = \cos(\theta/2)$. Then $x = (\sqrt{5} + 2)(1 - x)$ $\Rightarrow x = \sqrt{5} + 2 - x(\sqrt{5} + 2)$ $\Rightarrow x(1 + \sqrt{5} + 2) = \sqrt{5} + 2$ $\Rightarrow x = \frac{\sqrt{5} + 2}{\sqrt{5} + 3} = \frac{(\sqrt{5} + 2)(\sqrt{5} - 3)}{5 - 9} = \frac{5 - 3\sqrt{5} + 2\sqrt{5} - 6}{-4} = \frac{-1 - \sqrt{5}}{-4} = \frac{\sqrt{5} + 1}{4}$.
For $2\theta$,the ratio is $\frac{\cos \theta}{1 - \cos \theta}$.
Since $\cos \theta = 2\cos^2(\theta/2) - 1 = 2(\frac{\sqrt{5} + 1}{4})^2 - 1 = 2(\frac{5 + 1 + 2\sqrt{5}}{16}) - 1 = \frac{6 + 2\sqrt{5}}{8} - 1 = \frac{3 + \sqrt{5}}{4} - 1 = \frac{\sqrt{5} - 1}{4}$.
Ratio $= \frac{(\sqrt{5} - 1)/4}{1 - (\sqrt{5} - 1)/4} = \frac{\sqrt{5} - 1}{4 - \sqrt{5} + 1} = \frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

(D) Given $\angle AOB = \theta$ and radius $OF = OA = OB = d$.
In $\triangle ODB$,$BD = OB \sin \theta = d \sin \theta$.
Since $E$ is the mid-point of $BD$,$ED = \frac{1}{2} BD = \frac{d}{2} \sin \theta$.
Let $F$ be a point on the arc $AB$ such that $EF \parallel OA$. Let $FM \perp OA$ where $M$ is on $OA$. Then $FM = ED = \frac{d}{2} \sin \theta$.
In $\triangle OFM$,$\sin \alpha = \frac{FM}{OF} = \frac{\frac{d}{2} \sin \theta}{d} = \frac{1}{2} \sin \theta$,where $\alpha = \angle AOF$.
Thus,$\alpha = \sin^{-1}(\frac{1}{2} \sin \theta)$.
The length of arc $AF = d \alpha$ and the length of arc $AB = d \theta$.
The ratio of the length of arc $AF$ to the length of arc $AB$ is $\frac{d \alpha}{d \theta} = \frac{\alpha}{\theta} = \frac{\sin^{-1}(\frac{1}{2} \sin \theta)}{\theta}$.

(B) Let $O$ be the center of circle $C_1$ and $O'$ be the center of circle $C_2$. Let the line joining the centers be the $x$-axis.
Since the circles touch each other,the distance between their centers is $R-r$.
Let $M$ be the projection of $P$ onto the line $OO'$. Since $l$ is tangent to $C_1$ at $P$,$PM \perp OO'$,so $PM = r$.
Let $N$ be the projection of $B$ onto the line $OO'$. Since $l$ is parallel to $OO'$,$BN = PM = r$.
In $\triangle O'NB$,$O'B = R$ and $BN = r$.
Given $R^2 = 2r^2$,we have $R = \sqrt{2}r$.
Then $O'N = \sqrt{O'B^2 - BN^2} = \sqrt{2r^2 - r^2} = r$.
Since $O'N = BN = r$,$\triangle O'NB$ is an isosceles right-angled triangle,so $\angle BO'N = 45^{\circ}$.
Similarly,$\angle AO'M = 45^{\circ}$.
Thus,$\angle AO'B = 180^{\circ} - (45^{\circ} + 45^{\circ}) = 90^{\circ}$.
The angle subtended by the chord $AB$ at the center $O'$ is $90^{\circ}$.
The angle subtended by the same chord $AB$ at any point $O$ on the circumference of $C_2$ is half the angle subtended at the center.
Therefore,$\angle AOB = \frac{1}{2} \angle AO'B = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.

(B) Let $r$ be the radius of the circle. The center of the circle subtends angles at the center corresponding to the chords of lengths $1$ and $2$.
Let $2\theta$ be the angle subtended by a chord of length $1$ at the center,and $2\alpha$ be the angle subtended by a chord of length $2$ at the center.
Then,$\sin \theta = \frac{1/2}{r} = \frac{1}{2r}$ and $\sin \alpha = \frac{1}{r}$.
The sum of the angles around the center is $3(2\theta) + 3(2\alpha) = 360^{\circ}$,which implies $\theta + \alpha = 60^{\circ}$.
Taking cosine on both sides,$\cos(\theta + \alpha) = \cos(60^{\circ}) = \frac{1}{2}$.
Using the formula $\cos(\theta + \alpha) = \cos \theta \cos \alpha - \sin \theta \sin \alpha = \frac{1}{2}$.
Since $\sin \theta = \frac{1}{2r}$,$\cos \theta = \sqrt{1 - \frac{1}{4r^2}} = \frac{\sqrt{4r^2-1}}{2r}$.
Since $\sin \alpha = \frac{1}{r}$,$\cos \alpha = \sqrt{1 - \frac{1}{r^2}} = \frac{\sqrt{r^2-1}}{r}$.
Substituting these into the equation: $\frac{\sqrt{4r^2-1}}{2r} \cdot \frac{\sqrt{r^2-1}}{r} - \frac{1}{2r} \cdot \frac{1}{r} = \frac{1}{2}$.
$\frac{\sqrt{(4r^2-1)(r^2-1)}}{2r^2} = \frac{1}{2} + \frac{1}{2r^2} = \frac{r^2+1}{2r^2}$.
$\sqrt{(4r^2-1)(r^2-1)} = r^2+1$.
Squaring both sides: $(4r^2-1)(r^2-1) = (r^2+1)^2$.
$4r^4 - 5r^2 + 1 = r^4 + 2r^2 + 1$.
$3r^4 = 7r^2 \Rightarrow r^2 = \frac{7}{3}$.
Thus,$r = \sqrt{\frac{7}{3}}$.

(B) The equation of the circle $C$ is $x^2+y^2=1$.
The line $L_t$ passes through $(0,1)$ and $(t,0)$. Its equation is $\frac{x}{t} + \frac{y}{1} = 1$,which simplifies to $y = 1 - \frac{x}{t}$.
Substituting $y$ into the circle equation: $x^2 + (1 - \frac{x}{t})^2 = 1$.
$x^2 + 1 - \frac{2x}{t} + \frac{x^2}{t^2} = 1$.
$x^2(1 + \frac{1}{t^2}) = \frac{2x}{t}$.
$x^2(\frac{t^2+1}{t^2}) = \frac{2x}{t} \implies x = 0$ or $x = \frac{2t}{1+t^2}$.
The point $(0,1)$ corresponds to $x=0$. The other point $Q_t$ has $x = \frac{2t}{1+t^2}$.
Then $y = 1 - \frac{2}{1+t^2} = \frac{1+t^2-2}{1+t^2} = \frac{t^2-1}{t^2+1}$.
Let $t = \tan \theta$. Then $x = \sin 2\theta$ and $y = -\cos 2\theta$.
For $t \in [1, 1+\sqrt{2}]$,$\theta \in [\frac{\pi}{4}, \frac{3\pi}{8}]$.
The angle $2\theta$ varies from $\frac{\pi}{2}$ to $\frac{3\pi}{4}$.
The angle subtended by the arc at the origin is the difference in the polar angles of the endpoints: $\frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.