Geometrical problems regarding circle and its properties Questions in English

(C) Given that the radius $(r) = 3 \text{ m}$ and the arc length $(l) = 1 \text{ m}$.
We know that the formula for the angle $(\theta)$ subtended at the centre is $\theta = \frac{l}{r}$.
Substituting the values,we get $\theta = \frac{1}{3} \text{ radian}$.

(B) The circumference of the circular wire with radius $r_1 = 7\,cm$ is given by $L = 2\pi r_1 = 2 \times \pi \times 7 = 14\pi\,cm$.
This wire is bent into an arc of a circle with radius $r_2 = 12\,cm$. The length of the arc is $L = 14\pi\,cm$.
The formula for the angle $\theta$ subtended by an arc at the centre is $\theta = \frac{L}{r_2}$ (in radians).
$\theta = \frac{14\pi}{12} = \frac{7\pi}{6}$ radians.
To convert radians to degrees,multiply by $\frac{180^o}{\pi}$:
$\theta = \frac{7\pi}{6} \times \frac{180^o}{\pi} = 7 \times 30^o = 210^o$.

(B) The relationship between the arc length $(l)$,radius $(r)$,and the angle subtended at the centre in radians $( heta)$ is given by the formula: $l = r \theta$.
Given: Arc length $l = 15 \ cm$ and angle $\theta = 3/4 \ radian$.
Substituting the values into the formula: $15 = r \times (3/4)$.
Solving for $r$: $r = 15 \times (4/3) = 5 \times 4 = 20 \ cm$.
Therefore,the radius of the circle is $20 \ cm$.

(C) The circle is given by $x^2 + y^2 = 25$,which has its center at the origin $O(0, 0)$ and radius $r = 5$.
The coordinates of $Q$ are $(3, 4)$ and $R$ are $(-4, 3)$.
The slope of $OQ$ is $m_1 = \frac{4 - 0}{3 - 0} = \frac{4}{3}$.
The slope of $OR$ is $m_2 = \frac{3 - 0}{-4 - 0} = -\frac{3}{4}$.
Since $m_1 \times m_2 = \left(\frac{4}{3}\right) \times \left(-\frac{3}{4}\right) = -1$,the lines $OQ$ and $OR$ are perpendicular to each other.
Therefore,the central angle $\angle QOR = \frac{\pi}{2}$.
According to the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle QPR = \frac{1}{2} \angle QOR = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$.

(A) The given circle is $x^2 + y^2 - 6x - 4y - 12 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = -2, c = -12$.
The center of the circle is $(-g, -f) = (3, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-2)^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$.
Since the lines are parallel to the $x$-axis,their equations are of the form $y = k$.
The distance from the center $(3, 2)$ to the line $y - k = 0$ must be equal to the radius $r = 5$.
Using the perpendicular distance formula,$\frac{|2 - k|}{\sqrt{0^2 + 1^2}} = 5$.
$|2 - k| = 5$.
This gives $2 - k = 5$ or $2 - k = -5$.
So,$k = -3$ or $k = 7$.
The lines are $y = -3$ and $y = 7$,which can be written as $y + 3 = 0$ and $y - 7 = 0$.
The equation of the pair of lines is $(y + 3)(y - 7) = 0$.
$y^2 - 7y + 3y - 21 = 0$.
$y^2 - 4y - 21 = 0$.

(D) The equation of the circle is $x^2 + y^2 - 2x + 4y + 3 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 2$.
The center of the circle is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 3} = \sqrt{1 + 4 - 3} = \sqrt{2}$.
Since the sides of the square are parallel to the coordinate axes,the vertices of the square are at a distance $r$ from the center along the diagonals.
The vertices are $(1 \pm r \cos(45^\circ), -2 \pm r \sin(45^\circ)) = (1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, -2 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}) = (1 \pm 1, -2 \pm 1)$.
The vertices are $(2, -1), (0, -1), (0, -3), (2, -3)$.
Comparing these with the given options,none of the options match the calculated vertices.
Therefore,the correct option is $(d)$.

(D) The equation of the circle is $x^2 + y^2 - 6x + 2y = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -6 \Rightarrow g = -3$ and $2f = 2 \Rightarrow f = 1$.
The center of the circle is $(-g, -f) = (3, -1)$.
Since the line $x + 2by + 7 = 0$ is a diameter,the center of the circle must lie on this line.
Substituting $(3, -1)$ into the line equation: $3 + 2b(-1) + 7 = 0$.
$10 - 2b = 0$.
$2b = 10 \Rightarrow b = 5$.

(A) The radius of a circle that touches a line is equal to the perpendicular distance from the centre of the circle to that line.
The formula for the perpendicular distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Given the centre $(x_1, y_1) = (1, -3)$ and the line $3x - 4y - 5 = 0$,we have $A = 3$,$B = -4$,and $C = -5$.
Substituting these values into the formula:
$r = \frac{|3(1) - 4(-3) - 5|}{\sqrt{3^2 + (-4)^2}}$
$r = \frac{|3 + 12 - 5|}{\sqrt{9 + 16}}$
$r = \frac{|10|}{\sqrt{25}}$
$r = \frac{10}{5} = 2$.
Thus,the radius of the circle is $2$.

(C) The radius $r$ of the circle is the distance between the centre $(1, 2)$ and the point $(4, 6)$ on the circle.
Using the distance formula,$r = \sqrt{(4 - 1)^2 + (6 - 2)^2}$.
$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The area of the circle is given by the formula $A = \pi r^2$.
Substituting $r = 5$,we get $A = \pi (5)^2 = 25\pi \text{ sq. units}$.

(C) The given equation of the circle is $x^2 + y^2 + 4x - 4y + 4 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4 \implies g = 2$,$2f = -4 \implies f = -2$,and $c = 4$.
The centre of the circle is $(-g, -f) = (-2, 2)$.
The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + (-2)^2 - 4} = \sqrt{4 + 4 - 4} = \sqrt{4} = 2$.
Since the absolute value of the $x$-coordinate of the centre is $|-2| = 2 = r$ and the absolute value of the $y$-coordinate of the centre is $|2| = 2 = r$,the circle touches both the $x$-axis and the $y$-axis.

(B) The given circle is ${x^2} + {y^2} - 2x - 4y - 20 = 0$.
Its center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-20)} = \sqrt{25} = 5$.
Let the required circle have center $C_2 = (h, k)$ and radius $r_2 = 5$.
Since the circles touch externally at $(5, 5)$,the point $(5, 5)$ divides the line segment joining the centers $C_1(1, 2)$ and $C_2(h, k)$ in the ratio $r_1 : r_2 = 5 : 5 = 1 : 1$.
Thus,$(5, 5) = (\frac{1+h}{2}, \frac{2+k}{2})$.
Solving for $h$ and $k$:
$1 + h = 10 \Rightarrow h = 9$
$2 + k = 10 \Rightarrow k = 8$.
The equation of the required circle is $(x - 9)^2 + (y - 8)^2 = 5^2$.
Expanding this,we get ${x^2} - 18x + 81 + {y^2} - 16y + 64 = 25$.
${x^2} + {y^2} - 18x - 16y + 120 = 0$.

(B) Let the center of the circle be $(h, k)$.
Since the circle passes through $(-2, 0)$ and $(4, 0)$,the center must lie on the perpendicular bisector of the segment joining these points.
The midpoint of $(-2, 0)$ and $(4, 0)$ is $(\frac{-2+4}{2}, \frac{0+0}{2}) = (1, 0)$.
The perpendicular bisector is the vertical line $x = 1$,so $h = 1$.
The distance from the center $(1, k)$ to $(4, 0)$ is the radius $r = 5$.
Using the distance formula: $\sqrt{(4-1)^2 + (0-k)^2} = 5$.
$3^2 + k^2 = 5^2 \implies 9 + k^2 = 25$.
$k^2 = 16 \implies k = \pm 4$.
Thus,there are two possible centers: $(1, 4)$ and $(1, -4)$.
Therefore,there are $2$ such circles.

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.
Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.
Since the lines are parallel,the distance between them is the diameter of the circle.
The distance $d$ 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,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$.
$d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{|7.5|}{\sqrt{9 + 16}} = \frac{7.5}{5} = 1.5$.
Since the diameter is $1.5$,the radius $r = \frac{d}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$.

(C) Since the circle lies in the third quadrant and touches both axes,its centre is $(-a, -a)$ and its radius is $a$,where $a > 0$.
The equation of the circle is $(x + a)^2 + (y + a)^2 = a^2$,which simplifies to $x^2 + y^2 + 2ax + 2ay + a^2 = 0$.
The circle touches the line $3x - 4y + 8 = 0$. The perpendicular distance from the centre $(-a, -a)$ to the line must equal the radius $a$:
$\left| \frac{3(-a) - 4(-a) + 8}{\sqrt{3^2 + (-4)^2}} \right| = a$
$\left| \frac{-3a + 4a + 8}{5} \right| = a$
$\left| \frac{a + 8}{5} \right| = a$
Case $1$: $a + 8 = 5a$ $\Rightarrow 4a = 8$ $\Rightarrow a = 2$.
Case $2$: $a + 8 = -5a$ $\Rightarrow 6a = -8$ $\Rightarrow a = -4/3$ (rejected as $a > 0$).
Substituting $a = 2$ into the equation $x^2 + y^2 + 2ax + 2ay + a^2 = 0$ gives:
$x^2 + y^2 + 4x + 4y + 4 = 0$.

(C) The given equation of the circle is $x^2 + y^2 - 4x - 6y + 11 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -4 \implies g = -2$ and $2f = -6 \implies f = -3$.
The centre of the circle is $(-g, -f) = (2, 3)$.
Let the given end of the diameter be $A = (3, 4)$ and the other end be $B = (x, y)$.
Since the centre $C(2, 3)$ is the midpoint of the diameter $AB$,we have:
$\frac{3 + x}{2} = 2 \implies 3 + x = 4 \implies x = 1$
$\frac{4 + y}{2} = 3 \implies 4 + y = 6 \implies y = 2$
Thus,the other end is $(1, 2)$.

(B) The given equation of the circle is $x^2 + y^2 + 6x - 8y + 9 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 3$,$f = -4$,and $c = 9$.
The length of the intercept made by the circle on the $x$-axis is given by $2\sqrt{g^2 - c}$.
Substituting the values,we get $2\sqrt{3^2 - 9} = 2\sqrt{9 - 9} = 0$.
Since the intercept on the $x$-axis is $0$,the circle touches the $x$-axis.

(C) The equation of the circle is $x^2 + y^2 - 4x - 6y + 3 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$ and $f = -3$.
The centre of the circle is $(-g, -f) = (2, 3)$.
Now,check if the line $3x + 2y = 12$ passes through the centre $(2, 3)$ by substituting the coordinates:
$3(2) + 2(3) = 6 + 6 = 12$.
Since the line satisfies the equation,it passes through the centre of the circle.
Therefore,the line is a diameter of the circle.

(C) The given equation of the circle is $x^2 + y^2 - 6x - 8y - 9 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -6 \implies g = -3$ and $2f = -8 \implies f = -4$.
The centre of the circle is $(-g, -f) = (3, 4)$.
$A$ diameter of a circle always passes through its centre.
We check which option satisfies the point $(3, 4)$:
For option $C$: $x + y = 3 + 4 = 7$. This satisfies the equation.
Thus,the line $x + y = 7$ is a diameter.

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Its center is $(-g, -f)$ and its radius is $r = \sqrt{g^2 + f^2 - c}$.
For the circle to touch both axes,the distance from the center to both axes must be equal to the radius.
Thus,$|-g| = |-f| = r$,which implies $|g| = |f| = r$.
Also,the radius formula gives $r^2 = g^2 + f^2 - c$.
Substituting $g^2 = r^2$ and $f^2 = r^2$,we get $r^2 = r^2 + r^2 - c$,which simplifies to $c = r^2$.
Therefore,the condition is $g^2 = f^2 = c = r^2$.

(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
The center of the circle is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c} = a$.
Since the $x$-axis is a diameter,the center of the circle must lie on the $x$-axis.
Therefore,the $y$-coordinate of the center must be zero,which implies $-f = 0$,so $f = 0$.
Since the circle has the $x$-axis as a diameter,it passes through the origin if the center is on the $x$-axis and it touches the $x$-axis at the origin,but here it says $x$-axis is a diameter,meaning the circle intersects the $x$-axis at two points.
For the $x$-axis to be a diameter,the center $(-g, 0)$ must lie on the $x$-axis (which is true for any $f=0$) and the distance from the center to the $x$-axis must be $0$.
The radius is $a = \sqrt{g^2 + f^2 - c}$. Since $f = 0$,$a = \sqrt{g^2 - c}$.
Squaring both sides,$a^2 = g^2 - c$,which implies $c = g^2 - a^2$.
Looking at the options provided,none of the specific relations $f=2a, g=0, c=3a^2$ etc. are generally true for all such circles.
Thus,the correct choice is $(D)$.

(A) The given equation of the circle is $x^2 + y^2 - 6x + 2y = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -6 \implies g = -3$ and $2f = 2 \implies f = 1$.
The center of the circle is $(-g, -f) = (3, -1)$.
$A$ diameter of the circle always passes through its center.
We need the equation of a line passing through the origin $(0, 0)$ and the center $(3, -1)$.
The slope $m$ of the line is $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 0}{3 - 0} = -\frac{1}{3}$.
The equation of the line is $y - 0 = -\frac{1}{3}(x - 0)$,which simplifies to $3y = -x$ or $x + 3y = 0$.

(C) The given lines are $L_1: y + \sqrt{3}x = 6$,$L_2: y - \sqrt{3}x = 6$,and $L_3: y = 0$.
Finding the vertices of the triangle:
Intersection of $L_1$ and $L_2$: $y = 6, x = 0$. Vertex $A = (0, 6)$.
Intersection of $L_1$ and $L_3$: $y = 0, \sqrt{3}x = 6 \implies x = 2\sqrt{3}$. Vertex $B = (2\sqrt{3}, 0)$.
Intersection of $L_2$ and $L_3$: $y = 0, -\sqrt{3}x = 6 \implies x = -2\sqrt{3}$. Vertex $C = (-2\sqrt{3}, 0)$.
Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(0, 6)$,$(2\sqrt{3}, 0)$,and $(-2\sqrt{3}, 0)$:
For $(2\sqrt{3}, 0)$: $12 + 4g\sqrt{3} + c = 0$
For $(-2\sqrt{3}, 0)$: $12 - 4g\sqrt{3} + c = 0$
Adding these gives $24 + 2c = 0 \implies c = -12$. Then $4g\sqrt{3} = 0 \implies g = 0$.
For $(0, 6)$: $0^2 + 6^2 + 2f(6) - 12 = 0 \implies 36 + 12f - 12 = 0 \implies 12f = -24 \implies f = -2$.
The equation is $x^2 + y^2 - 4y - 12 = 0$,which is $x^2 + y^2 - 4y = 12$.

(C) The area of a circle is given by the formula $A = \pi r^2$,where $r$ is the radius of the circle.
Given that the radius is $a$,we substitute $r = a$ into the formula.
Therefore,the area is $A = \pi a^2$.

(C) Let $AB$ be the chord of length $\sqrt{2}$ and $O$ be the centre of the circle.
Given that the angle subtended by the chord at the centre is $\angle AOB = \frac{\pi}{2} = 90^\circ$.
In $\Delta AOB$,$OA = OB = r$ (radius of the circle).
By the Pythagorean theorem in $\Delta AOB$:
$OA^2 + OB^2 = AB^2$
$r^2 + r^2 = (\sqrt{2})^2$
$2r^2 = 2$
$r^2 = 1$
Therefore,the area of the circle is $\pi r^2 = \pi(1) = \pi$.

(D) The given equation of the circle is $x^2 + y^2 - 8x - 4y + c = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -8$ and $2f = -4$,which implies $g = -4$ and $f = -2$.
The center of the circle is $(-g, -f) = (4, 2)$.
Let the coordinates of the other end of the diameter be $(x, y)$.
Since the center is the midpoint of the diameter,we have:
$\frac{x + (-3)}{2} = 4 \implies x - 3 = 8 \implies x = 11$
$\frac{y + 2}{2} = 2 \implies y + 2 = 4 \implies y = 2$
Thus,the coordinates of the other end are $(11, 2)$.

(A) The centre of a circle is the midpoint of its diameter.
Given the endpoints of the diameter are $(x, 3)$ and $(3, 5)$,and the centre is $(2, y)$.
Using the midpoint formula,the centre is $(\frac{x + 3}{2}, \frac{3 + 5}{2})$.
Equating the coordinates with the given centre $(2, y)$:
$\frac{x + 3}{2} = 2 \implies x + 3 = 4 \implies x = 1$.
$\frac{3 + 5}{2} = y \implies \frac{8}{2} = y \implies y = 4$.
Thus,$x = 1$ and $y = 4$.

(B) The radius of a circle passing through two fixed points is minimum when the line segment joining the two points is the diameter of the circle.
Given points are $A(1, 0)$ and $B(0, 1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the points $(1, 0)$ and $(0, 1)$:
$(x - 1)(x - 0) + (y - 0)(y - 1) = 0$
$x(x - 1) + y(y - 1) = 0$
$x^2 - x + y^2 - y = 0$
$x^2 + y^2 - x - y = 0$
Thus,the correct option is $B$.

(A) The given equation is $x^2 + y^2 = 2ax$.
Rearranging the terms,we get $x^2 - 2ax + y^2 = 0$.
Completing the square for $x$,we add $a^2$ to both sides: $(x^2 - 2ax + a^2) + y^2 = a^2$.
This simplifies to $(x - a)^2 + y^2 = a^2$.
This is the equation of a circle with center $(a, 0)$ and radius $r = a$.
The area of a circle is given by the formula $A = \pi r^2$.
Substituting $r = a$,we get $A = \pi a^2$.

(B) Let $r$ be the radius of the circle and $a$ be the side length of the square.
Given that the perimeter of the circle equals the perimeter of the square:
$2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$.
Area of the circle = $\pi r^2$.
Area of the square = $a^2 = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4}$.
Comparing the areas:
$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi}$.
Since $\pi \approx 3.14$,$\frac{4}{\pi} > 1$.
Therefore,the area of the circle is larger than the area of the square.

(D) Comparing the given equation $x^2 + y^2 + 10x - 6y + 9 = 0$ with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 5$ and $c = 9$.
The length of the intercept made by the circle on the $x$-axis is given by the formula $2\sqrt{g^2 - c}$.
Substituting the values,we get $2\sqrt{5^2 - 9} = 2\sqrt{25 - 9} = 2\sqrt{16} = 2 \times 4 = 8$.
Therefore,the length of the intercept is $8$.

(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 + 3x - 3y + 2 = 0$.
$A$ point $(x_1, y_1)$ lies inside the circle if $S(x_1, y_1) < 0$.
Checking option $(A) (-1, 3)$: $S(-1, 3) = (-1)^2 + (3)^2 + 3(-1) - 3(3) + 2 = 1 + 9 - 3 - 9 + 2 = 0$. The point lies on the circle.
Checking option $(B) (-2, 1)$: $S(-2, 1) = (-2)^2 + (1)^2 + 3(-2) - 3(1) + 2 = 4 + 1 - 6 - 3 + 2 = -2$. Since $-2 < 0$,the point lies inside the circle.
Checking option $(C) (2, 1)$: $S(2, 1) = (2)^2 + (1)^2 + 3(2) - 3(1) + 2 = 4 + 1 + 6 - 3 + 2 = 10$. Since $10 > 0$,the point lies outside the circle.
Checking option $(D) (-3, 2)$: $S(-3, 2) = (-3)^2 + (2)^2 + 3(-3) - 3(2) + 2 = 9 + 4 - 9 - 6 + 2 = 0$. The point lies on the circle.
Thus,the correct option is $(B)$.

(A) Let the equation of the circle be $S(x, y) = x^2 + y^2 - x + y - 1 = 0$.
To find the position of the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the expression $S(x, y)$.
$S(1, 1) = (1)^2 + (1)^2 - (1) + (1) - 1 = 1 + 1 - 1 + 1 - 1 = 1$.
Since $S(1, 1) = 1 > 0$,the point $(1, 1)$ lies outside the circle.

(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{S_1}$,where $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Given the point $(5, 1)$ and the circle $x^2 + y^2 + 6x - 4y - 3 = 0$.
Substituting the point into the circle equation:
$S_1 = (5)^2 + (1)^2 + 6(5) - 4(1) - 3$
$S_1 = 25 + 1 + 30 - 4 - 3 = 49$
Therefore,the length of the tangent is $\sqrt{S_1} = \sqrt{49} = 7$.

(C) The given circle is $x^2 + y^2 - 4x - 8y - 5 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$.
The centre of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$.
For the line $3x - 4y - m = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre $(2, 4)$ to the line must be less than the radius $r$.
Distance $d = \frac{|3(2) - 4(4) - m|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - m|}{\sqrt{9 + 16}} = \frac{|-10 - m|}{5}$.
We require $d < r$,so $\frac{|-10 - m|}{5} < 5$.
$|-10 - m| < 25$.
$-25 < -10 - m < 25$.
Multiplying by $-1$,we get $25 > 10 + m > -25$.
Subtracting $10$ from all parts,we get $15 > m > -35$,which is $-35 < m < 15$.

(C) The condition for the line $y = mx + c$ to be a tangent to the circle ${x^2} + {y^2} = {a^2}$ is ${c^2} = {a^2}(1 + {m^2})$.
If ${c^2} < {a^2}(1 + {m^2})$,the line intersects the circle at two distinct points.
If ${c^2} > {a^2}(1 + {m^2})$,the distance from the center $(0, 0)$ to the line $mx - y + c = 0$ is given by $d = \frac{|c|}{\sqrt{m^2 + 1}}$.
Since ${c^2} > {a^2}(1 + {m^2})$,we have $\frac{|c|}{\sqrt{m^2 + 1}} > |a|$.
This means the perpendicular distance from the center to the line is greater than the radius of the circle.
Therefore,the line does not intersect the circle at any point.

(A) Since the circle touches both coordinate axes and lies in the first quadrant,its center is $(c, c)$ and its radius is $r = c$.
The distance from the center $(c, c)$ to the line $\frac{x}{3} + \frac{y}{4} = 1$ (or $4x + 3y - 12 = 0$) must be equal to the radius $c$.
Using the perpendicular distance formula: $\left| \frac{4c + 3c - 12}{\sqrt{4^2 + 3^2}} \right| = c$.
$\left| \frac{7c - 12}{5} \right| = c$.
This gives two cases:
Case $1$: $7c - 12 = 5c$ $\Rightarrow 2c = 12$ $\Rightarrow c = 6$.
Case $2$: $7c - 12 = -5c$ $\Rightarrow 12c = 12$ $\Rightarrow c = 1$.
Since the line $\frac{x}{3} + \frac{y}{4} = 1$ has intercepts $3$ and $4$,the circle must be smaller than the triangle formed by the axes and the line. For $c=6$,the circle lies outside the triangle. Thus,$c=1$ is the correct value for the circle touching the line internally.

(D) The equation of the circle is $x^2 + y^2 + 2x - 4y + 4 = 0$.
Let the point be $P(-1, 2)$.
To check the position of the point with respect to the circle,we calculate the power of the point $S_1$ by substituting the coordinates into the circle equation:
$S_1 = (-1)^2 + (2)^2 + 2(-1) - 4(2) + 4$
$S_1 = 1 + 4 - 2 - 8 + 4 = -1$.
Since $S_1 < 0$,the point $(-1, 2)$ lies inside the circle.
Therefore,no tangent can be drawn from this point to the circle.

(A) Let the equation of the circle be $S: x^2 + y^2 + 2x + 6y - 15 = 0$.
To determine the position of the point $(0, 0)$ with respect to the circle,we substitute the coordinates $(0, 0)$ into the expression $S$.
$S_1 = (0)^2 + (0)^2 + 2(0) + 6(0) - 15 = -15$.
Since $S_1 < 0$,the point $(0, 0)$ lies inside the circle.
Therefore,no tangent can be drawn from a point inside the circle to the circle itself.

(D) Let the equation of the circle be $S(x, y) = x^2 + y^2 - 2x - 4y + 3 = 0$.
To determine the position of the point $(0.1, 3.1)$,we calculate the power of the point $S_1 = S(0.1, 3.1)$.
$S_1 = (0.1)^2 + (3.1)^2 - 2(0.1) - 4(3.1) + 3$
$S_1 = 0.01 + 9.61 - 0.2 - 12.4 + 3$
$S_1 = 12.62 - 12.6 = 0.02$.
Since $S_1 > 0$,the point lies outside the circle.

(A) Given line: $4x - 3y - 10 = 0 \implies x = \frac{3y + 10}{4}$.
Substituting this into the circle equation $x^2 + y^2 - 2x + 4y - 20 = 0$:
$(\frac{3y + 10}{4})^2 + y^2 - 2(\frac{3y + 10}{4}) + 4y - 20 = 0$.
Multiplying by $16$ to clear the denominator:
$(3y + 10)^2 + 16y^2 - 8(3y + 10) + 64y - 320 = 0$.
$9y^2 + 60y + 100 + 16y^2 - 24y - 80 + 64y - 320 = 0$.
$25y^2 + 100y - 300 = 0$.
$y^2 + 4y - 12 = 0$.
$(y + 6)(y - 2) = 0$.
So,$y = -6$ or $y = 2$.
For $y = -6$,$x = \frac{3(-6) + 10}{4} = \frac{-8}{4} = -2$.
For $y = 2$,$x = \frac{3(2) + 10}{4} = \frac{16}{4} = 4$.
The points are $(-2, -6)$ and $(4, 2)$.

(C) The perpendicular distance from the center of the circle $(0, 0)$ to the line $mx - y + c = 0$ must be less than the radius $r$ for the line to intersect the circle at two distinct points.
The distance $d$ is given by $d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}}$.
For two distinct points,we require $d < r$,so $\frac{|c|}{\sqrt{1 + m^2}} < r$.
This implies $|c| < r\sqrt{1 + m^2}$,which is equivalent to $c^2 < r^2(1 + m^2)$.
This inequality can also be written as $-r\sqrt{1 + m^2} < c < r\sqrt{1 + m^2}$.

(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = 2$.
For the second circle $x^2 + y^2 - 8x + 12 = 0$,we rewrite it as $(x - 4)^2 + y^2 = 4$,so the center $C_2 = (4, 0)$ and radius $r_2 = 2$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4 - 0)^2 + (0 - 0)^2} = 4$.
Since $d = r_1 + r_2$ $(4 = 2 + 2)$,the two circles touch each other externally.
When two circles touch externally,the number of common tangents is $3$.

(D) The given circles are ${x^2} + {y^2} - x = 0$ and ${x^2} + {y^2} + x = 0$.
For the first circle ${x^2} + {y^2} - x = 0$,the centre ${C_1} = (\frac{1}{2}, 0)$ and radius ${r_1} = \sqrt{(\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.
For the second circle ${x^2} + {y^2} + x = 0$,the centre ${C_2} = (-\frac{1}{2}, 0)$ and radius ${r_2} = \sqrt{(-\frac{1}{2})^2 + 0^2 - 0} = \frac{1}{2}$.
The distance between the centres is ${C_1}{C_2} = \sqrt{(\frac{1}{2} - (-\frac{1}{2}))^2 + (0 - 0)^2} = \sqrt{1^2} = 1$.
Since ${C_1}{C_2} = 1$ and ${r_1} + {r_2} = \frac{1}{2} + \frac{1}{2} = 1$,we have ${C_1}{C_2} = {r_1} + {r_2}$.
This indicates that the two circles touch each other externally.
When two circles touch each other externally,the number of common tangents is $3$.

(B) The circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$.
This implies that the normal to the curve $y = x^2 + 1$ at $(1, 2)$ must pass through the center of the circle $(h, k)$.
The derivative of the curve is $dy/dx = 2x$. At $x = 1$,the slope of the tangent is $m_t = 2(1) = 2$.
The slope of the normal at $(1, 2)$ is $m_n = -1/m_t = -1/2$.
The equation of the normal line passing through $(1, 2)$ with slope $-1/2$ is $(y - 2) = -1/2(x - 1)$,which simplifies to $2y - 4 = -x + 1$,or $x + 2y = 5$.
Since the center $(h, k)$ lies on this normal line,we have $h + 2k = 5$.

(C) normal to a circle is a line that passes through the center of the circle.
Since the given line $ax + by + c = 0$ is a normal to the circle $x^2 + y^2 = r^2$,it must pass through the center of the circle,which is $(0, 0)$.
Any line passing through the center of a circle is a diameter of that circle.
The length of the diameter of a circle with radius $r$ is $2r$.
Therefore,the length of the portion of the line intercepted by the circle is $2r$.

(C) For the first circle $x^2 + y^2 = 1$,the center $C_1 = (0, 0)$ and radius $r_1 = 1$.
For the second circle $x^2 + y^2 - 4x + 3 = 0$,we rewrite it as $(x-2)^2 + y^2 = 1$,so the center $C_2 = (2, 0)$ and radius $r_2 = 1$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2-0)^2 + (0-0)^2} = 2$.
Since $d = r_1 + r_2 = 1 + 1 = 2$,the two circles touch each other externally.
When two circles touch externally,the number of common tangents is $3$.

(C) The equation of a chord of a circle $S = 0$ with mid-point $(x_1, y_1)$ is given by $T = S_1$,where $T = x x_1 + y y_1 - a^2$ and $S_1 = x_1^2 + y_1^2 - a^2$.
Substituting these into the formula:
$x x_1 + y y_1 - a^2 = x_1^2 + y_1^2 - a^2$
Adding $a^2$ to both sides,we get:
$x x_1 + y y_1 = x_1^2 + y_1^2$.

(B) The length of the chord is given by the formula $L = 2\sqrt{R^2 - d^2}$,where $R$ is the radius of the circle and $d$ is the perpendicular distance from the center $(0, 0)$ to the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$.
The radius $R = r$.
The perpendicular distance $d = \frac{|\frac{0}{a} + \frac{0}{b} - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{a^2 + b^2}{a^2b^2}}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.
Substituting these into the formula:
$L = 2\sqrt{r^2 - \frac{a^2b^2}{a^2 + b^2}}$
$L = 2\sqrt{\frac{r^2(a^2 + b^2) - a^2b^2}{a^2 + b^2}}$.

(C) Let the midpoint of the chord be $M(h, k)$.
Since the line $x - 2y = 2$ is the chord,the midpoint $M$ must lie on this line,so $h - 2k = 2$.
The line segment joining the center of the circle $(0, 0)$ to the midpoint $M(h, k)$ is perpendicular to the chord $x - 2y = 2$.
The slope of the chord is $m_1 = \frac{1}{2}$.
The slope of the line segment $OM$ is $m_2 = \frac{k - 0}{h - 0} = \frac{k}{h}$.
Since $OM$ is perpendicular to the chord,$m_1 \times m_2 = -1$,which gives $\frac{1}{2} \times \frac{k}{h} = -1$,so $k = -2h$.
Substituting $k = -2h$ into the equation $h - 2k = 2$,we get $h - 2(-2h) = 2$,which implies $5h = 2$,so $h = \frac{2}{5}$.
Then $k = -2 \times \frac{2}{5} = -\frac{4}{5}$.
Thus,the midpoint is $\left( \frac{2}{5}, -\frac{4}{5} \right)$.

(C) The given circle is $x^2 + y^2 + 2x - 4y - 11 = 0$. The center of this circle is $(-g, -f) = (-1, 2)$.
The diameter that bisects a chord is perpendicular to that chord.
The given line is $2x - y + 3 = 0$,which has a slope of $m_1 = 2$.
The slope of the diameter perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$.
The equation of the line passing through the center $(-1, 2)$ with slope $-\frac{1}{2}$ is given by $y - 2 = -\frac{1}{2}(x + 1)$.
Multiplying by $2$,we get $2y - 4 = -x - 1$.
Rearranging the terms,we get $x + 2y - 3 = 0$.