Equations of circle Questions in English

(A) circle that touches both axes with radius $a$ has its center at $(a, a)$,$(a, -a)$,$(-a, a)$,or $(-a, -a)$.
Considering the case where the center is $(a, a)$,the equation is $(x - a)^2 + (y - a)^2 = a^2$.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$.
Simplifying,we obtain $x^2 + y^2 - 2ax - 2ay + a^2 = 0$.

(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the centre is $(-g, -f)$.
For the first circle $x^2 + y^2 - 1 = 0$,the centre $C_1$ is $(0, 0)$.
For the second circle $x^2 + y^2 + 6x - 2y - 1 = 0$,$2g = 6 \implies g = 3$ and $2f = -2 \implies f = -1$. The centre $C_2$ is $(-3, 1)$.
For the third circle $x^2 + y^2 - 12x + 4y - 1 = 0$,$2g = -12 \implies g = -6$ and $2f = 4 \implies f = 2$. The centre $C_3$ is $(6, -2)$.
To check if $C_1(0, 0)$,$C_2(-3, 1)$,and $C_3(6, -2)$ are collinear,we check the slope between pairs.
Slope of $C_1C_2 = \frac{1 - 0}{-3 - 0} = -\frac{1}{3}$.
Slope of $C_2C_3 = \frac{-2 - 1}{6 - (-3)} = \frac{-3}{9} = -\frac{1}{3}$.
Since the slopes are equal,the points are collinear.

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(0, 0)$,we have $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$,which gives $c = 0$.
Since it passes through $(a, 0)$,we have $a^2 + 0^2 + 2g(a) + 2f(0) + 0 = 0$,which implies $a^2 + 2ga = 0$,so $g = -\frac{a}{2}$.
Since it passes through $(0, b)$,we have $0^2 + b^2 + 2g(0) + 2f(b) + 0 = 0$,which implies $b^2 + 2fb = 0$,so $f = -\frac{b}{2}$.
The centre of the circle is $(-g, -f)$.
Substituting the values of $g$ and $f$,the centre is $\left( -(-\frac{a}{2}), -(-\frac{b}{2}) \right) = \left( \frac{a}{2}, \frac{b}{2} \right)$.

(A) The radius $r$ of the circle is the perpendicular distance from the center $(1, -3)$ to the line $2x - y - 4 = 0$.
$r = \left| \frac{2(1) - (-3) - 4}{\sqrt{2^2 + (-1)^2}} \right| = \left| \frac{2 + 3 - 4}{\sqrt{5}} \right| = \frac{1}{\sqrt{5}}$.
The equation of the circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 1)^2 + (y + 3)^2 = \left( \frac{1}{\sqrt{5}} \right)^2$.
$(x^2 - 2x + 1) + (y^2 + 6y + 9) = \frac{1}{5}$.
$x^2 + y^2 - 2x + 6y + 10 = \frac{1}{5}$.
Multiplying by $5$,we get $5x^2 + 5y^2 - 10x + 30y + 50 = 1$.
$5x^2 + 5y^2 - 10x + 30y + 49 = 0$.

(B) The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Given the centre is $(x_1, y_1)$ and the circle touches both axes,the radius $r$ must satisfy $r = |x_1| = |y_1|$.
Thus,$x_1 = y_1 = r$ (assuming the circle is in the first quadrant,$x_1^2 = y_1^2 = r^2$).
Substituting these into the equation: $(x - x_1)^2 + (y - x_1)^2 = x_1^2$.
Expanding the terms: $(x^2 - 2x x_1 + x_1^2) + (y^2 - 2y x_1 + x_1^2) = x_1^2$.
Simplifying: $x^2 + y^2 - 2x_1(x + y) + x_1^2 = 0$.

(B) The center of the circle is the point of intersection of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.
Solving these equations: $2x - 3y = 5$ (multiplied by $3$) $\Rightarrow 6x - 9y = 15$ and $3x - 4y = 7$ (multiplied by $2$) $\Rightarrow 6x - 8y = 14$.
Subtracting the first from the second: $(6x - 8y) - (6x - 9y) = 14 - 15 \Rightarrow y = -1$.
Substituting $y = -1$ into $2x - 3y = 5$: $2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.
So,the center $(h, k) = (1, -1)$.
Given area $= 154$,we have $\pi r^2 = 154$ $\Rightarrow \frac{22}{7} r^2 = 154$ $\Rightarrow r^2 = 154 \times \frac{7}{22} = 49$ $\Rightarrow r = 7$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 7^2$.
$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
$x^2 + y^2 - 2x + 2y + 2 = 49$.
$x^2 + y^2 - 2x + 2y = 47$.

(C) Let the center of the circle be $O' = (h, k)$.
Since the circle touches the $y$-axis at $(0, 4)$,the $y$-coordinate of the center is $k = 4$ and the radius $r = |h|$.
Thus,the equation of the circle is $(x - h)^2 + (y - 4)^2 = h^2$.
This circle cuts the $x$-axis $(y = 0)$ at points where $(x - h)^2 + (0 - 4)^2 = h^2$,which simplifies to $(x - h)^2 + 16 = h^2$,or $(x - h)^2 = h^2 - 16$.
So,$x - h = \pm \sqrt{h^2 - 16}$,which gives $x = h \pm \sqrt{h^2 - 16}$.
The length of the chord on the $x$-axis is the difference between these $x$-values: $(h + \sqrt{h^2 - 16}) - (h - \sqrt{h^2 - 16}) = 2\sqrt{h^2 - 16}$.
Given the chord length is $6$,we have $2\sqrt{h^2 - 16} = 6$,so $\sqrt{h^2 - 16} = 3$.
Squaring both sides,$h^2 - 16 = 9$,which gives $h^2 = 25$,so $h = 5$.
Therefore,the radius $r = |h| = 5$.

(B) The centre of the circle is $(h, k) = (1, 2)$.
Since the circle touches the $x$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre,so $r = |2| = 2$.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 1)^2 + (y - 2)^2 = 2^2$.
Expanding this,we get $(x^2 - 2x + 1) + (y^2 - 4y + 4) = 4$.
Simplifying,we get $x^2 + y^2 - 2x - 4y + 1 = 0$.

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the given equation $x^2 + y^2 - 18x + 12y + k = 0$ with the general form,we get $2g = -18 \implies g = -9$ and $2f = 12 \implies f = 6$.
The constant term is $c = k$.
The radius $r$ of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.
Given $r = 11$,we have $11 = \sqrt{(-9)^2 + (6)^2 - k}$.
Squaring both sides,$121 = 81 + 36 - k$.
$121 = 117 - k$.
$k = 117 - 121 = -4$.

(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
The centre of the circle is the midpoint of the diameter.
Therefore,the centre is $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.

(B) Since $\angle C = 90^{\circ}$,the hypotenuse $AB$ must be the diameter of the circumcircle of $\triangle ABC$ because the angle subtended by the diameter at the circumference is a right angle.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the coordinates $A(-3, 4)$ and $B(3, -4)$:
$(x - (-3))(x - 3) + (y - 4)(y - (-4)) = 0$
$(x + 3)(x - 3) + (y - 4)(y + 4) = 0$
$(x^2 - 9) + (y^2 - 16) = 0$
$x^2 + y^2 - 25 = 0$
$x^2 + y^2 = 25$.

(A) Since the circle is in the first quadrant and touches both coordinate axes at a distance of $1$ unit from the origin,the center of the circle is $(1, 1)$ and the radius $r = 1$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $h = 1, k = 1$,and $r = 1$:
$(x - 1)^2 + (y - 1)^2 = 1^2$
$x^2 - 2x + 1 + y^2 - 2y + 1 = 1$
$x^2 + y^2 - 2x - 2y + 1 = 0$.

(B) Let the general equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(2, -2)$,$(-1, -1)$,and $(5, 2)$,we have:
$1) \; 4 + 4 + 4g - 4f + c = 0 \Rightarrow 4g - 4f + c = -8$
$2) \; 1 + 1 - 2g - 2f + c = 0 \Rightarrow -2g - 2f + c = -2$
$3) \; 25 + 4 + 10g + 4f + c = 0 \Rightarrow 10g + 4f + c = -29$
Subtracting $(2)$ from $(1)$: $6g - 2f = -6$ $\Rightarrow 3g - f = -3$ $\Rightarrow f = 3g + 3$.
Subtracting $(2)$ from $(3)$: $12g + 6f = -27 \Rightarrow 4g + 2f = -9$.
Substituting $f = 3g + 3$ into $4g + 2f = -9$: $4g + 2(3g + 3) = -9$ $\Rightarrow 10g + 6 = -9$ $\Rightarrow 10g = -15$ $\Rightarrow g = -1.5$.
Then $f = 3(-1.5) + 3 = -4.5 + 3 = -1.5$.
Substituting $g$ and $f$ into $(2)$: $-2(-1.5) - 2(-1.5) + c = -2$ $\Rightarrow 3 + 3 + c = -2$ $\Rightarrow c = -8$.
The equation is $x^2 + y^2 - 3x - 3y - 8 = 0$.

(D) Since the circle passes through the origin $(0, 0)$ and cuts intercepts of length $3$ and $4$ on the positive $x$ and $y$ axes respectively,it passes through the points $(3, 0)$ and $(0, 4)$.
The general equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.
Substituting the point $(3, 0)$ into the equation:
$3^2 + 0^2 + 2g(3) + 2f(0) = 0$ $\Rightarrow 9 + 6g = 0$ $\Rightarrow g = -\frac{3}{2}$.
Substituting the point $(0, 4)$ into the equation:
$0^2 + 4^2 + 2g(0) + 2f(4) = 0$ $\Rightarrow 16 + 8f = 0$ $\Rightarrow f = -2$.
Substituting $g$ and $f$ back into the general equation:
$x^2 + y^2 + 2(-\frac{3}{2})x + 2(-2)y = 0$
$x^2 + y^2 - 3x - 4y = 0$.

(B) The given equation of the circle is $x^2 + y^2 + 6y = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = 3$,and $c = 0$.
The center of the circle is $(-g, -f) = (0, -3)$.
The radius of the circle is $R = \sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 3^2 - 0} = 3$.
Since the center is $(0, -3)$ and the radius is $3$,the distance from the center to the $x$-axis is $|-3| = 3$,which is equal to the radius.
Therefore,the circle touches the $x$-axis at the point $(0, 0)$,which is the origin.

(A) The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$.
The radius $R$ of this circle is given by the formula $R = \sqrt{g^2 + f^2 - c}$.
$A$ circle is defined as a point circle when its radius is equal to zero.
Setting the radius to zero,we get $\sqrt{g^2 + f^2 - c} = 0$.
Squaring both sides,we obtain $g^2 + f^2 - c = 0$,which simplifies to $g^2 + f^2 = c$.

(A) First,find the point of intersection of the lines $3x + y = 14$ and $2x + 5y = 18$.
Multiplying the first equation by $5$,we get $15x + 5y = 70$.
Subtracting the second equation from this,we get $(15x - 2x) = 70 - 18$,which gives $13x = 52$,so $x = 4$.
Substituting $x = 4$ into $3x + y = 14$,we get $3(4) + y = 14$,so $y = 2$.
The point of intersection is $(4, 2)$.
The radius $r$ is the distance between the center $(1, -2)$ and the point $(4, 2)$:
$r^2 = (4 - 1)^2 + (2 - (-2))^2 = 3^2 + 4^2 = 9 + 16 = 25$.
The equation of the circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $(h, k) = (1, -2)$ and $r^2 = 25$,we get $(x - 1)^2 + (y + 2)^2 = 25$.
Expanding this,$x^2 - 2x + 1 + y^2 + 4y + 4 = 25$,which simplifies to $x^2 + y^2 - 2x + 4y - 20 = 0$.

(B) Let the centre of the circle be $(h, k)$ and radius be $r$.
Since the circle touches the lines $x = 0$ and $y = 0$,the centre must be at $(r, r)$ or $(r, -r)$ or $(-r, r)$ or $(-r, -r)$.
Assuming the circle is in the first quadrant,the centre is $(r, r)$ and the equation is $(x - r)^2 + (y - r)^2 = r^2$.
This circle also touches the line $3x + 4y - 4 = 0$.
The perpendicular distance from the centre $(r, r)$ to the line is equal to the radius $r$:
$\frac{|3r + 4r - 4|}{\sqrt{3^2 + 4^2}} = r$
$\frac{|7r - 4|}{5} = r$
$|7r - 4| = 5r$
Case $1$: $7r - 4 = 5r$ $\Rightarrow 2r = 4$ $\Rightarrow r = 2$.
The equation is $(x - 2)^2 + (y - 2)^2 = 2^2$,which simplifies to $x^2 + y^2 - 4x - 4y + 4 = 0$.
Case $2$: $7r - 4 = -5r$ $\Rightarrow 12r = 4$ $\Rightarrow r = 1/3$.
Comparing with the given options,the correct equation is $x^2 - 4x + y^2 - 4y + 4 = 0$.

(B) The centre of the circle is $(h, k) = (2, 2)$.
Since the circle passes through $(4, 5)$,the radius $r$ is the distance between $(2, 2)$ and $(4, 5)$:
$r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values: $(x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2$.
$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 13$.
$x^2 + y^2 - 4x - 4y + 8 = 13$.
$x^2 + y^2 - 4x - 4y - 5 = 0$.

(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Comparing this with the given equation $(x - 5)(x - 1) + (y - 7)(y - 4) = 0$,the endpoints of the diameter are $(5, 7)$ and $(1, 4)$.
The length of the diameter $d$ is the distance between these two points:
$d = \sqrt{(5 - 1)^2 + (7 - 4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
The radius $r$ is half of the diameter:
$r = \frac{d}{2} = \frac{5}{2}$.

(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$ and $(4, 5)$,the distances from the centre to these points are equal (equal to the radius $r$):
$(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$
$4h + 4k = 28 \implies h + k = 7$
Given that the centre lies on $y - 4x + 3 = 0$,we have $k - 4h + 3 = 0$.
Solving $h + k = 7$ and $k - 4h = -3$:
Subtracting the equations: $(h + k) - (k - 4h) = 7 - (-3) \implies 5h = 10 \implies h = 2$.
Then $k = 7 - 2 = 5$.
The centre is $(2, 5)$.
The radius squared is $r^2 = (2 - 2)^2 + (5 - 3)^2 = 0^2 + 2^2 = 4$.
The equation of the circle is $(x - 2)^2 + (y - 5)^2 = 4$.
$x^2 - 4x + 4 + y^2 - 10y + 25 = 4$
$x^2 + y^2 - 4x - 10y + 25 = 0$.

(A) The given circle is $x^2 + y^2 + 2y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$ and $f = 1$. The centre of this circle is $(-g, -f) = (0, -1)$.
The required circle has its centre at $(1, -2)$ and passes through $(0, -1)$.
The radius $r$ is the distance between $(1, -2)$ and $(0, -1)$,which is $r = \sqrt{(1 - 0)^2 + (-2 - (-1))^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k) = (1, -2)$.
$(x - 1)^2 + (y + 2)^2 = (\sqrt{2})^2$
$x^2 - 2x + 1 + y^2 + 4y + 4 = 2$
$x^2 + y^2 - 2x + 4y + 3 = 0$.

(B) The given circle is $x^2 + y^2 + 8x + 10y - 7 = 0$. Any circle concentric with this circle is of the form $x^2 + y^2 + 8x + 10y + k = 0$.
The centre of the circle $x^2 + y^2 - 4x - 6y = 0$ is found by comparing it with $x^2 + y^2 + 2gx + 2fy + c = 0$,which gives $(-g, -f) = (2, 3)$.
Since the required circle passes through $(2, 3)$,we substitute these coordinates into the equation:
$(2)^2 + (3)^2 + 8(2) + 10(3) + k = 0$
$4 + 9 + 16 + 30 + k = 0$
$59 + k = 0 \implies k = -59$.
Thus,the equation of the circle is $x^2 + y^2 + 8x + 10y - 59 = 0$.

(C) The general equation of a circle passing through the origin $(0, 0)$ is given by $x^2 + y^2 + 2gx + 2fy = 0$ ... $(i)$.
Since the circle passes through $(0, b)$,we substitute $x = 0$ and $y = b$ into $(i)$:
$0^2 + b^2 + 2g(0) + 2f(b) = 0$ $\Rightarrow b^2 + 2fb = 0$ $\Rightarrow f = -\frac{b}{2}$.
Since the circle passes through $(a, b)$,we substitute $x = a$ and $y = b$ into $(i)$:
$a^2 + b^2 + 2g(a) + 2f(b) = 0$.
Substituting $f = -\frac{b}{2}$ into this equation:
$a^2 + b^2 + 2ag + 2(-\frac{b}{2})(b) = 0$ $\Rightarrow a^2 + b^2 + 2ag - b^2 = 0$ $\Rightarrow a^2 + 2ag = 0$ $\Rightarrow g = -\frac{a}{2}$.
Substituting $g = -\frac{a}{2}$ and $f = -\frac{b}{2}$ back into $(i)$:
$x^2 + y^2 + 2(-\frac{a}{2})x + 2(-\frac{b}{2})y = 0 \Rightarrow x^2 + y^2 - ax - by = 0$.

(C) The general equation of a second-degree curve is given by $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$.
For this equation to represent a circle,the following conditions must be satisfied:
$1$. The coefficient of $x^2$ must be equal to the coefficient of $y^2$,i.e.,$a = b$.
$2$. The coefficient of $xy$ must be zero,i.e.,$h = 0$.
$3$. The coefficients $a$ and $b$ must not be zero $(a = b \neq 0)$.
Therefore,the correct condition is $a = b \neq 0$ and $h = 0$.

(A) The equation of a circle touching both axes in the first quadrant is $(x - a)^2 + (y - a)^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax - 2ay + a^2 = 0$.
Since the circle passes through the point $(1, 2)$,we substitute these coordinates into the equation:
$(1 - a)^2 + (2 - a)^2 = a^2$
$1 - 2a + a^2 + 4 - 4a + a^2 = a^2$
$a^2 - 6a + 5 = 0$
$(a - 1)(a - 5) = 0$
Thus,$a = 1$ or $a = 5$.
For $a = 1$,the equation is $x^2 + y^2 - 2x - 2y + 1 = 0$.
For $a = 5$,the equation is $x^2 + y^2 - 10x - 10y + 25 = 0$.

(A) The given circle is $x^2 + y^2 - 6x + 12y + 15 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $g = -3, f = 6, c = 15$.
The radius $r_1$ of the given circle is $\sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 6^2 - 15} = \sqrt{9 + 36 - 15} = \sqrt{30}$.
The area of the given circle is $A_1 = \pi r_1^2 = 30\pi$.
The required circle is concentric,so its equation is $x^2 + y^2 - 6x + 12y + k = 0$.
Its radius $r_2$ satisfies $r_2^2 = g^2 + f^2 - k = 45 - k$.
The area of the required circle is $A_2 = \pi r_2^2 = \pi(45 - k)$.
Given that $A_2 = 2A_1$,we have $\pi(45 - k) = 2(30\pi) = 60\pi$.
$45 - k = 60 \Rightarrow k = -15$.
Thus,the equation is $x^2 + y^2 - 6x + 12y - 15 = 0$.

(C) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis.
Given the radius $r = 4$ and the circle passes through the origin $(0, 0)$.
The distance from the centre $(h, 0)$ to the origin $(0, 0)$ must be equal to the radius $4$.
So,$\sqrt{(h-0)^2 + (0-0)^2} = 4$,which implies $|h| = 4$,so $h = \pm 4$.
The centre is $(\pm 4, 0)$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting $h = \pm 4, k = 0, r = 4$:
$(x \mp 4)^2 + (y-0)^2 = 4^2$
$x^2 \mp 8x + 16 + y^2 = 16$
$x^2 + y^2 \mp 8x = 0$.

(B) Since the circle touches the $y$-axis at the origin $(0, 0)$,its center must lie on the $x$-axis. Let the center be $(h, 0)$.
Since it touches the $y$-axis at the origin,the radius of the circle is $|h|$.
The equation of the circle is $(x - h)^2 + (y - 0)^2 = h^2$,which simplifies to ${x^2} - 2hx + {h^2} + {y^2} = {h^2}$,or ${x^2} + {y^2} - 2hx = 0$.
Since the circle passes through the point $(2, 1)$,we substitute these coordinates into the equation:
${2^2} + {1^2} - 2h(2) = 0$
$4 + 1 - 4h = 0$
$5 - 4h = 0$
$h = \frac{5}{4}$
Substituting $h = \frac{5}{4}$ back into the equation ${x^2} + {y^2} - 2hx = 0$:
${x^2} + {y^2} - 2(\frac{5}{4})x = 0$
${x^2} + {y^2} - \frac{5}{2}x = 0$
Multiplying by $2$,we get $2{x^2} + 2{y^2} - 5x = 0$.
Thus,the correct option is $B$.

(C) Since the circle passes through the origin $(0, 0)$,the constant term $c = 0$.
The circle cuts off intercepts of length $2$ on the negative coordinate axes.
The length of the intercept on the $x$-axis is $2\sqrt{g^2 - c} = 2$. Since the intercept is on the negative axis,the center's $x$-coordinate $-g$ must be negative,so $g = 1$.
The length of the intercept on the $y$-axis is $2\sqrt{f^2 - c} = 2$. Since the intercept is on the negative axis,the center's $y$-coordinate $-f$ must be negative,so $f = 1$.
The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting $g = 1$,$f = 1$,and $c = 0$,we get $x^2 + y^2 + 2x + 2y = 0$.

(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the given equation $x^2 + y^2 + 3x + 3y = 0$ with the general form,we get $c = 0$.
If the constant term $c = 0$,the circle passes through the origin $(0, 0)$.
Therefore,the correct option is $D$.

(A) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis. The radius is $r = 5$.
The equation of the circle is $(x - h)^2 + (y - 0)^2 = 5^2$,which simplifies to $(x - h)^2 + y^2 = 25$.
Since the circle passes through the point $(2, 3)$,we substitute these coordinates into the equation:
$(2 - h)^2 + 3^2 = 25$
$(2 - h)^2 + 9 = 25$
$(2 - h)^2 = 16$
$2 - h = \pm 4$
Case $1$: $2 - h = 4 \Rightarrow h = -2$. The equation is $(x + 2)^2 + y^2 = 25$ $\Rightarrow x^2 + 4x + 4 + y^2 = 25$ $\Rightarrow x^2 + y^2 + 4x - 21 = 0$.
Case $2$: $2 - h = -4 \Rightarrow h = 6$. The equation is $(x - 6)^2 + y^2 = 25$ $\Rightarrow x^2 - 12x + 36 + y^2 = 25$ $\Rightarrow x^2 + y^2 - 12x + 11 = 0$.
Comparing with the given options,the correct equation is $x^2 + y^2 + 4x - 21 = 0$.

(A) Since the circle touches the $x$-axis at $(3, 0)$,its center is $(3, k)$ and its radius is $|k|$.
The equation of the circle is $(x - 3)^2 + (y - k)^2 = k^2$.
Since it passes through $(1, 4)$,we substitute these coordinates into the equation:
$(1 - 3)^2 + (4 - k)^2 = k^2$
$(-2)^2 + 16 - 8k + k^2 = k^2$
$4 + 16 - 8k = 0$
$20 = 8k$
$k = \frac{20}{8} = \frac{5}{2}$.
Substituting $k = \frac{5}{2}$ back into the equation:
$(x - 3)^2 + (y - \frac{5}{2})^2 = (\frac{5}{2})^2$
$x^2 - 6x + 9 + y^2 - 5y + \frac{25}{4} = \frac{25}{4}$
$x^2 + y^2 - 6x - 5y + 9 = 0$.

(A) Given that the diameter of the circle is $20$,the radius $r = \frac{20}{2} = 10$.
The center of the circle is the point of intersection of the two diameters $x + y = 6$ and $x + 2y = 4$.
Subtracting the first equation from the second: $(x + 2y) - (x + y) = 4 - 6 \Rightarrow y = -2$.
Substituting $y = -2$ into $x + y = 6$,we get $x - 2 = 6 \Rightarrow x = 8$.
So,the center of the circle is $(h, k) = (8, -2)$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 8)^2 + (y + 2)^2 = 10^2$.
$x^2 - 16x + 64 + y^2 + 4y + 4 = 100$.
$x^2 + y^2 - 16x + 4y + 68 - 100 = 0$.
$x^2 + y^2 - 16x + 4y - 32 = 0$.

(B) Let the circle have center $(h, k)$ and radius $r$.
Since the circle touches the lines $y = a$ and $y = b$,the diameter of the circle must be equal to the distance between these two parallel lines.
Thus,$2r = |b - a|$,which implies $r = \frac{|b - a|}{2}$.
The distance from the center $(h, k)$ to the line $y = a$ is $|k - a| = r$,so $k = a \pm r$.
The distance from the center $(h, k)$ to the line $x = 0$ is $|h| = r$,so $h = \pm r$.
Since $r$ is fixed,for each value of $h$ ($r$ or $-r$),we get a circle. Thus,there are exactly $2$ such circles.

(A) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the given points $(a, 0)$ and $(0, b)$:
$(x - a)(x - 0) + (y - 0)(y - b) = 0$
$x(x - a) + y(y - b) = 0$
$x^2 - ax + y^2 - by = 0$
$x^2 + y^2 - ax - by = 0$.

(A) The given equation of the circle is $2x^2 + 2y^2 - x = 0$.
Dividing the entire equation by $2$,we get $x^2 + y^2 - \frac{1}{2}x = 0$.
Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{1}{2} \implies g = -\frac{1}{4}$,$2f = 0 \implies f = 0$,and $c = 0$.
The centre of the circle is $(-g, -f) = \left( \frac{1}{4}, 0 \right)$.
The radius $R$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{\left( -\frac{1}{4} \right)^2 + 0^2 - 0} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the centre is $\left( \frac{1}{4}, 0 \right)$ and the radius is $\frac{1}{4}$.

(A) The standard equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre of the circle and $r$ is the radius.
Comparing the given equation $(x - 3)^2 + (y - 4)^2 = 5$ with the standard form:
$h = 3$ and $k = 4$.
Therefore,the centre of the circle is $(3, 4)$.

(D) The circle touches the lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $x = 4$.
Since it touches $x = 0$ and $x = 4$,the diameter of the circle is the distance between these lines,which is $4 - 0 = 4$. Thus,the radius $r = 2$.
The center of the circle must be at a distance $r = 2$ from both $x = 0$ and $y = 0$,so the center is $(2, 2)$.
The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $h = 2, k = 2, r = 2$,we get $(x - 2)^2 + (y - 2)^2 = 2^2$.
$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 4$.
$x^2 + y^2 - 4x - 4y + 4 = 0$.

(A) The equation $x^2 + y^2 = 0$ represents a circle with center $(0, 0)$ and radius $r = \sqrt{0^2 + 0^2 - 0} = 0$.
Since the radius is $0$,the circle degenerates into a single point,which is the origin $(0, 0)$.

(D) For a general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a circle,the following conditions must be met:
$1$. The coefficient of $x^2$ must equal the coefficient of $y^2$,so $a = 2$.
$2$. The coefficient of $xy$ must be zero,so $2b = 0$,which implies $b = 0$.
$3$. Since the circle passes through the origin $(0, 0)$,the constant term $c$ must be $0$.
Thus,the values are $a = 2, b = 0, c = 0$.

(C) The distance between the lines $x = 1$ and $x = -1$ is $|1 - (-1)| = |1 + 1| = 2$.
Thus,the radius of the circle is $r = 2$.
The centre of the circle is the origin $(0, 0)$.
The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 0)^2 + (y - 0)^2 = 2^2$.
Therefore,the equation of the circle is $x^2 + y^2 = 4$.

(A) Since the circle touches the $x$-axis at $(3, 0)$,the $x$-coordinate of the centre is $3$.
Since the circle touches the $y$-axis at $(0, -3)$,the $y$-coordinate of the centre is $-3$.
Thus,the centre of the circle is $(3, -3)$.

(A) The radius $r$ of the circle is the perpendicular distance from the centre $(2, 3)$ to the tangent line $x + y - 1 = 0$.
Using the formula for perpendicular distance $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we get:
$r = \frac{|1(2) + 1(3) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The square of the radius is $r^2 = (2\sqrt{2})^2 = 8$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting $(h, k) = (2, 3)$ and $r^2 = 8$,we get $(x - 2)^2 + (y - 3)^2 = 8$.

(A) Since the circle passes through the origin $(0, 0)$ and cuts intercepts $a$ and $b$ on the $x$ and $y$ axes respectively,the circle passes through the points $(a, 0)$ and $(0, b)$.
Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it passes through $(0, 0)$,we have $c = 0$.
Since it passes through $(a, 0)$,we have $a^2 + 2ga = 0$,which implies $g = -a/2$.
Since it passes through $(0, b)$,we have $b^2 + 2fb = 0$,which implies $f = -b/2$.
Substituting these values into the general equation,we get $x^2 + y^2 - ax - by = 0$.

(A) Two circles are concentric if they have the same center.
The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
The reference circle is $x^2 + y^2 - 2x + 4y + 20 = 0$.
Comparing the reference circle with $x^2 + y^2 + 2gx + 2fy + c = 0$,the center is $(-g, -f)$.
For the reference circle,the center is $(-(-1), -(2)) = (1, -2)$.
Thus,for the required circle,$g = -1$ and $f = 2$.
The equation becomes $x^2 + y^2 - 2x + 4y + c = 0$.
Since the circle passes through $(4, -2)$,we substitute these coordinates into the equation:
$(4)^2 + (-2)^2 - 2(4) + 4(-2) + c = 0$
$16 + 4 - 8 - 8 + c = 0$
$4 + c = 0$
$c = -4$.

(C) Let the center of the circle be $(h, k)$.
Since the circle touches the $y$-axis at $(0,3)$,the $y$-coordinate of the center is $k = 3$ and the radius $r = |h|$.
The equation of the circle is $(x - h)^2 + (y - 3)^2 = h^2$.
This circle cuts an intercept of $8$ units on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^2 + (0 - 3)^2 = h^2$,which simplifies to $(x - h)^2 + 9 = h^2$,or $(x - h)^2 = h^2 - 9$.
Thus,$x = h \pm \sqrt{h^2 - 9}$.
The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^2 - 9}) - (h - \sqrt{h^2 - 9}) = 2\sqrt{h^2 - 9}$.
Given the intercept is $8$,we have $2\sqrt{h^2 - 9} = 8$,so $\sqrt{h^2 - 9} = 4$.
Squaring both sides gives $h^2 - 9 = 16$,so $h^2 = 25$,which means $h = \pm 5$.
Since the radius $r = |h|$,we have $r = 5$.

(A) The centre of the given circle is $C_1 = (-4, 3)$ and its radius is $r_1$. The circle $x^2 + y^2 = 1$ has centre $C_2 = (0, 0)$ and radius $r_2 = 1$.
The distance between the centres is $d = \sqrt{(-4 - 0)^2 + (3 - 0)^2} = \sqrt{16 + 9} = 5$.
If the circles touch internally,$r_1 = d + r_2 = 5 + 1 = 6$ or $r_1 = |d - r_2| = |5 - 1| = 4$.
Case $1$: If $r_1 = 4$,the equation is $(x + 4)^2 + (y - 3)^2 = 4^2 \implies x^2 + 8x + 16 + y^2 - 6y + 9 = 16 \implies x^2 + y^2 + 8x - 6y + 9 = 0$.
Case $2$: If $r_1 = 6$,the equation is $(x + 4)^2 + (y - 3)^2 = 6^2 \implies x^2 + 8x + 16 + y^2 - 6y + 9 = 36 \implies x^2 + y^2 + 8x - 6y - 11 = 0$.
Comparing with the options,option $A$ matches the result from Case $1$.

(B) The equation of a circle concentric with the given circle $x^2 + y^2 - 4x - 6y - 3 = 0$ is of the form $x^2 + y^2 - 4x - 6y + k = 0$.
The center of this circle is $(2, 3)$.
Since the circle touches the $y$-axis $(x = 0)$,the radius of the circle is equal to the absolute value of the $x$-coordinate of the center.
Radius $r = |2| = 2$.
The formula for the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$.
Here,$g = -2$,$f = -3$,and $c = k$.
So,$\sqrt{(-2)^2 + (-3)^2 - k} = 2$.
Squaring both sides,we get $4 + 9 - k = 4$.
$13 - k = 4 \Rightarrow k = 9$.
Substituting $k = 9$ into the equation,we get $x^2 + y^2 - 4x - 6y + 9 = 0$.