Equations of circle Questions in English

(A) The circle passes through points $(4, 0)$ and $(-4, 0)$ on the $x$-axis.
Since the points are symmetric about the $y$-axis,the center of the circle must lie on the $y$-axis.
Let the center be $(0, c)$.
The distance from the center $(0, c)$ to the point $(4, 0)$ is the radius $r = 5$.
Using the distance formula: $\sqrt{(4-0)^2 + (0-c)^2} = 5$.
$16 + c^2 = 25$ $\Rightarrow c^2 = 9$ $\Rightarrow c = \pm 3$.
Thus,the centers are $(0, 3)$ and $(0, -3)$.
For center $(0, 3)$,the equation is $x^2 + (y-3)^2 = 5^2$ $\Rightarrow x^2 + y^2 - 6y + 9 = 25$ $\Rightarrow x^2 + y^2 - 6y - 16 = 0$.
For center $(0, -3)$,the equation is $x^2 + (y+3)^2 = 5^2$ $\Rightarrow x^2 + y^2 + 6y + 9 = 25$ $\Rightarrow x^2 + y^2 + 6y - 16 = 0$.
Both $x^2 + y^2 - 6y - 16 = 0$ and $x^2 + y^2 + 6y - 16 = 0$ are valid equations.

(D) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through the origin $(0,0)$,we have $c=0$.
The circle cuts off an intercept of $-2$ on the $x$-axis,meaning it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.
The circle cuts off an intercept of $3$ on the $y$-axis,meaning it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.
Substituting $g=1$,$f=-\frac{3}{2}$,and $c=0$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.
This simplifies to $x^2+y^2+2x-3y=0$.

(D) Given the equation of the circle: $x^2+y^2-3x-4y+2=0$.
Since the circle cuts the $x$-axis,the $y$-coordinate at these points must be $0$.
Substituting $y=0$ into the equation:
$x^2 + (0)^2 - 3x - 4(0) + 2 = 0$
$x^2 - 3x + 2 = 0$
Factoring the quadratic equation:
$(x-1)(x-2) = 0$
This gives $x=1$ and $x=2$.
Therefore,the points of intersection are $(1,0)$ and $(2,0)$.

(A) The given equation of the circle is $x^2+y^2+8x+10y-8=0$.
Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get:
$2g = 8 \implies g = 4$
$2f = 10 \implies f = 5$
$c = -8$
The centre of the circle is $(-g, -f) = (-4, -5)$.
The radius $r$ is given by $\sqrt{g^2+f^2-c}$.
$r = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$.
Thus,the centre is $(-4, -5)$ and the radius is $7$.

(D) The end points of the diameter are $(4,7)$ and $(-2,-1)$.
The equation of the circle is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the values,we get $(x-4)(x+2) + (y-7)(y+1) = 0$.
Expanding this,we have $x^2 - 2x - 8 + y^2 - 6y - 7 = 0$,which simplifies to $x^2 + y^2 - 2x - 6y - 15 = 0$.
The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the equations,we get $g = -1$,$f = -3$,and $c = -15$.
The length of the $X$-intercept is given by the formula $2\sqrt{g^2 - c}$.
Substituting the values,$AB = 2\sqrt{(-1)^2 - (-15)} = 2\sqrt{1 + 15} = 2\sqrt{16} = 2 \times 4 = 8$.
Thus,$AB = 8$.

(B) The circumference of the circle is $10 \pi$. Given $2 \pi r = 10 \pi$,we find the radius $r = 5$.
Since the diameters of a circle intersect at its center,we solve the system of equations:
$2x + 3y + 1 = 0$ $(i)$
$3x - y - 4 = 0$ $(ii)$
From $(ii)$,$y = 3x - 4$. Substituting this into $(i)$:
$2x + 3(3x - 4) + 1 = 0$
$2x + 9x - 12 + 1 = 0$
$11x - 11 = 0 \Rightarrow x = 1$.
Then $y = 3(1) - 4 = -1$.
The center of the circle is $(1, -1)$.
The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 5^2$
$x^2 - 2x + 1 + y^2 + 2y + 1 = 25$
$x^2 + y^2 - 2x + 2y - 23 = 0$.

(B) The circumference of the circle is $4 \pi$.
Since the circumference is given by $2 \pi r = 4 \pi$,we have $r = 2$.
The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the center $(-2, 3)$ and radius $r = 2$:
$(x - (-2))^2 + (y - 3)^2 = 2^2$
$(x + 2)^2 + (y - 3)^2 = 4$
$x^2 + 4x + 4 + y^2 - 6y + 9 = 4$
$x^2 + y^2 + 4x - 6y + 9 = 0$.

(B) The center of the circle is the point of intersection of its diameters.
Given equations of diameters:
$x + 2y - 5 = 0$ ... $(i)$
$3x - y - 1 = 0$ ... $(ii)$
Multiplying $(ii)$ by $2$,we get $6x - 2y - 2 = 0$ ... $(iii)$
Adding $(i)$ and $(iii)$:
$(x + 2y - 5) + (6x - 2y - 2) = 0$
$7x - 7 = 0 \Rightarrow x = 1$
Substituting $x = 1$ in $(i)$:
$1 + 2y - 5 = 0$ $\Rightarrow 2y = 4$ $\Rightarrow y = 2$
So,the center $(h, k) = (1, 2)$ and the radius $r = 5$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y - 2)^2 = 5^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = 25$
$x^2 + y^2 - 2x - 4y + 5 - 25 = 0$
$x^2 + y^2 - 2x - 4y - 20 = 0$

(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$.
Since the circle touches the $Y$-axis at $(0, 1)$,the $y$-coordinate of the center is $1$.
Thus,the center of the circle is $(1, 1)$ and the radius is $r = 1$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values,we get $(x-1)^2 + (y-1)^2 = 1^2$.
Expanding this,we have $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$.
Simplifying,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.

(D) The centre of the circle is $(h, k) = (5, 4)$.
Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the centre.
Thus,$r = |5| = 5$.
The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values,we get $(x-5)^2 + (y-4)^2 = 5^2$.
Expanding this,we have $(x^2 - 10x + 25) + (y^2 - 8y + 16) = 25$.
Simplifying,$x^2 + y^2 - 10x - 8y + 41 = 25$.
Therefore,$x^2 + y^2 - 10x - 8y + 16 = 0$.

(C) The given equation of the circle is $x^2+y^2+6x+2ky+25=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$,$f=k$,and $c=25$.
The center of the circle is $(-g, -f) = (-3, -k)$ and the radius $r$ is given by $\sqrt{g^2+f^2-c} = \sqrt{3^2+k^2-25} = \sqrt{k^2-16}$.
Since the circle touches the $Y$-axis,the radius must be equal to the absolute value of the $x$-coordinate of the center,i.e.,$r = |-g| = |-3| = 3$.
Equating the two expressions for the radius: $\sqrt{k^2-16} = 3$.
Squaring both sides: $k^2-16 = 9$.
$k^2 = 25$,which gives $k = \pm 5$.

(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.
Therefore,the circle passes through the points $(a,0)$ and $(0,b)$.
Since the angle between the coordinate axes is $90^{\circ}$,the line segment joining $(a,0)$ and $(0,b)$ is a diameter of the circle.
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 $(x_1, y_1) = (a,0)$ and $(x_2, y_2) = (0,b)$:
$(x-a)(x-0) + (y-0)(y-b) = 0$
$x(x-a) + y(y-b) = 0$
$x^2 - ax + y^2 - by = 0$
$x^2 + y^2 - ax - by = 0$

(B) The given equation of the circle is $x^2+y^2-6x-8y=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$,$f=-4$,and $c=0$.
The radius $r$ of the circle is given by the formula $r = \sqrt{g^2+f^2-c}$.
Substituting the values,$r = \sqrt{(-3)^2+(-4)^2-0} = \sqrt{9+16} = \sqrt{25} = 5 \text{ units}$.
The diameter of the circle is $d = 2r = 2 \times 5 = 10 \text{ units}$.

(D) The point of reflection of $(m, n)$ on the line $y-x=0$ is $(n, m)$.
Let the centre of the circle be $(n, m)$ and the radius be $r$.
The equation of the circle is $(x-n)^2 + (y-m)^2 = r^2$ ... $(i)$.
Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre,so $r = |m|$.
Thus,$r^2 = m^2$.
Substituting $r^2$ into equation $(i)$:
$(x-n)^2 + (y-m)^2 = m^2$
$x^2 - 2nx + n^2 + y^2 - 2my + m^2 = m^2$
$x^2 + y^2 - 2nx - 2my + n^2 = 0$.

(D) Centre $C = (2,3)$.
Radius $r$ is the perpendicular distance from the centre $(2,3)$ to the line $3x-4y+1=0$.
$r = \frac{|3(2)-4(3)+1|}{\sqrt{3^2+(-4)^2}} = \frac{|6-12+1|}{\sqrt{9+16}} = \frac{|-5|}{5} = 1$.
The equation of the circle is $(x-h)^2+(y-k)^2 = r^2$.
$(x-2)^2+(y-3)^2 = 1^2$.
$x^2-4x+4+y^2-6y+9 = 1$.
$x^2+y^2-4x-6y+12 = 0$.
Thus,option $D$ is correct.

(A) The centre of the circle $x^2+y^2-3x-4y-1=0$ is $C(\frac{3}{2}, 2)$.
Since the required circle passes through the point $C(\frac{3}{2}, 2)$ and has its centre at $(5, 2)$,the radius $r$ of the required circle is the distance between these two points:
$r = \sqrt{(5 - \frac{3}{2})^2 + (2 - 2)^2} = \sqrt{(\frac{7}{2})^2} = \frac{7}{2}$.
The equation of the required circle is $(x-5)^2 + (y-2)^2 = r^2$.
$(x-5)^2 + (y-2)^2 = (\frac{7}{2})^2$
$x^2 - 10x + 25 + y^2 - 4y + 4 = \frac{49}{4}$
$x^2 + y^2 - 10x - 4y + 29 = \frac{49}{4}$
Multiplying by $4$ on both sides:
$4x^2 + 4y^2 - 40x - 16y + 116 = 49$
$4x^2 + 4y^2 - 40x - 16y + 67 = 0$.
Thus,option $A$ is correct.

(A) The given equation of the circle is $(x+1)(x+2)+(y-1)(y+3)=0$.
Expanding the terms,we get $x^2+3x+2+y^2+2y-3=0$,which simplifies to $x^2+y^2+3x+2y-1=0$.
Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $2g=3 \implies g=\frac{3}{2}$,$2f=2 \implies f=1$,and $c=-1$.
The radius $r$ is given by $r=\sqrt{g^2+f^2-c} = \sqrt{(\frac{3}{2})^2 + (1)^2 - (-1)} = \sqrt{\frac{9}{4}+1+1} = \sqrt{\frac{9+8}{4}} = \sqrt{\frac{17}{4}}$.
The area of the circle is $\pi r^2 = \pi (\frac{17}{4}) = \frac{17 \pi}{4}$.
Thus,option $A$ is correct.

(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 $(h, k) = (1, 2)$.
Since the circle touches the $x$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre.
$r = |k| = |2| = 2$.
Substituting $h=1$,$k=2$,and $r=2$ into the standard equation:
$(x-1)^2+(y-2)^2=2^2$
$(x-1)^2+(y-2)^2=4$.

(B) Given equation: $2 x^2 + 2 y^2 - 3 x + 2 y - 1 = 0$
Divide by $2$: $x^2 + y^2 - \frac{3}{2} x + y - \frac{1}{2} = 0$
Comparing with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get:
$2g = -\frac{3}{2} \Rightarrow g = -\frac{3}{4}$
$2f = 1 \Rightarrow f = \frac{1}{2}$
$c = -\frac{1}{2}$
The radius $r$ is given by $r = \sqrt{g^2 + f^2 - c}$
$r = \sqrt{(-\frac{3}{4})^2 + (\frac{1}{2})^2 - (-\frac{1}{2})}$
$r = \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}}$
$r = \sqrt{\frac{9 + 4 + 8}{16}} = \sqrt{\frac{21}{16}}$
$r = \frac{\sqrt{21}}{4} \text{ units.}$

(C) The general equation of a circle passing through the origin $(0,0)$ is given by $x^2+y^2+2gx+2fy=0$.
Since the circle cuts off intercepts on the axes,the points of intersection are $(5,0)$ and $(0,6)$.
Substituting $(5,0)$ into the equation: $5^2+0^2+2g(5)+2f(0)=0 \implies 25+10g=0 \implies g = -2.5$.
Substituting $(0,6)$ into the equation: $0^2+6^2+2g(0)+2f(6)=0 \implies 36+12f=0 \implies f = -3$.
Substituting $g$ and $f$ back into the general equation: $x^2+y^2+2(-2.5)x+2(-3)y=0$.
This simplifies to $x^2+y^2-5x-6y=0$.

(A) Let the coordinates of points $P$ and $Q$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.
Given that $x_1$ and $x_2$ are the roots of $2x^2 + 4x - 7 = 0$,the sum of the roots is $x_1 + x_2 = -\frac{4}{2} = -2$.
Given that $y_1$ and $y_2$ are the roots of $3x^2 - 12x - 1 = 0$,the sum of the roots is $y_1 + y_2 = -\frac{-12}{3} = 4$.
The centre of the circle with $PQ$ as a diameter is the midpoint of $PQ$,which is given by $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.
Substituting the values,we get $(\frac{-2}{2}, \frac{4}{2}) = (-1, 2)$.
Thus,the correct option is $A$.

(A) Given: The lines $x+2y-5=0$ and $2x-3y+4=0$ are diameters of the circle.
Since the intersection of diameters is the center of the circle,we solve the system of equations:
$x+2y=5$ $(i)$
$2x-3y=-4$ (ii)
Multiplying $(i)$ by $2$,we get $2x+4y=10$ (iii).
Subtracting (ii) from (iii): $(2x+4y)-(2x-3y) = 10-(-4)$ $\Rightarrow 7y=14$ $\Rightarrow y=2$.
Substituting $y=2$ into $(i)$: $x+2(2)=5 \Rightarrow x=1$.
Thus,the center $(h, k)$ is $(1, 2)$.
The area of the circle is $9\pi$,so $\pi r^2 = 9\pi \Rightarrow r^2=9$.
The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$.
$(x-1)^2 + (y-2)^2 = 9
(x^2-2x+1) + (y^2-4y+4) = 9
x^2+y^2-2x-4y-4=0$.

(B) The given equation of the circle is $x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0$.
Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we get $g=\lambda$ and $f=-\lambda$.
The centre is $(-g, -f) = (-\lambda, \lambda)$.
Since the circle is in the second quadrant and touches the coordinate axes,the radius $r = |\lambda| = \lambda$ (as $\lambda > 0$ for the second quadrant).
The line is $\frac{x}{5}+\frac{y}{12}=1$,which simplifies to $12x+5y-60=0$.
The perpendicular distance from the centre $(-\lambda, \lambda)$ to the line $12x+5y-60=0$ must be equal to the radius $\lambda$.
$r = \frac{|12(-\lambda)+5(\lambda)-60|}{\sqrt{12^2+5^2}} = \lambda$.
$\frac{|-7\lambda-60|}{13} = \lambda$.
$|-7\lambda-60| = 13\lambda$.
This gives two cases:
$1) -7\lambda-60 = 13\lambda$ $\Rightarrow 20\lambda = -60$ $\Rightarrow \lambda = -3$.
$2) -7\lambda-60 = -13\lambda$ $\Rightarrow 6\lambda = 60$ $\Rightarrow \lambda = 10$.
Since the centre is $(-\lambda, \lambda)$ and it lies in the second quadrant,$\lambda$ must be positive.
Therefore,$\lambda = 10$.

(A) The distance between the parallel lines $3x - 4y - 10 = 0$ and $3x - 4y + 30 = 0$ is the diameter of the circle.
Diameter $d = \frac{|30 - (-10)|}{\sqrt{3^2 + (-4)^2}} = \frac{40}{5} = 8$.
Thus,the radius $r = \frac{d}{2} = 4$.
The centre $(h, k)$ lies on the line $x + 2y = 0$,so $h = -2k$.
The centre is equidistant from the two parallel lines. The line $3x - 4y + c = 0$ midway between the given lines is $3x - 4y + 10 = 0$.
Substituting $x = -2k$ into $3x - 4y + 10 = 0$ gives $3(-2k) - 4k + 10 = 0$,which simplifies to $-10k = -10$,so $k = 1$.
Then $h = -2(1) = -2$. The centre is $(-2, 1)$.
The equation of the circle is $(x + 2)^2 + (y - 1)^2 = 4^2$.
$x^2 + 4x + 4 + y^2 - 2y + 1 = 16$.
$x^2 + y^2 + 4x - 2y - 11 = 0$.

(B) The vertices of the square are $A(0,0)$,$B(16,0)$,$C(16,16)$,and $D(0,16)$.
Since the circle circumscribes the square,the diagonal $AC$ is the diameter of the circle.
The coordinates of $A$ are $(0,0)$ and $C$ are $(16,16)$.
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 coordinates of $A$ and $C$:
$(x-0)(x-16) + (y-0)(y-16) = 0$
$x(x-16) + y(y-16) = 0$
$x^2 - 16x + y^2 - 16y = 0$
$x^2 + y^2 = 16(x+y)$
Given the equation $x^2 + y^2 = 4k(x+y)$,we compare the two equations:
$4k = 16$
$k = 4$

(C) The roots of $x^2+5x+6=0$ are $\alpha, \beta$. By Vieta's formulas,$\alpha+\beta = -5$ and $\alpha\beta = 6$.
The roots of $y^2+6y+7=0$ are $\gamma, \delta$. By Vieta's formulas,$\gamma+\delta = -6$ and $\gamma\delta = 7$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Here,the endpoints are $(\alpha, \gamma)$ and $(\beta, \delta)$.
Thus,the equation is $(x-\alpha)(x-\beta) + (y-\gamma)(y-\delta) = 0$.
Expanding this,we get $x^2 - (\alpha+\beta)x + \alpha\beta + y^2 - (\gamma+\delta)y + \gamma\delta = 0$.
Substituting the values: $x^2 - (-5)x + 6 + y^2 - (-6)y + 7 = 0$.
This simplifies to $x^2 + y^2 + 5x + 6y + 13 = 0$.

(A) The sides of the rectangle are $x=4, x=-2, y=5, y=-2$.
The vertices of the rectangle are the intersection points of these lines: $(4, 5), (-2, 5), (-2, -2),$ and $(4, -2)$.
The diagonal of the rectangle connects opposite vertices,for example,$(4, 5)$ and $(-2, -2)$.
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 $(4, 5)$ and $(-2, -2)$:
$(x-4)(x+2) + (y-5)(y+2) = 0$
$x^2 + 2x - 4x - 8 + y^2 + 2y - 5y - 10 = 0$
$x^2 + y^2 - 2x - 3y - 18 = 0$.

(D) The equation of the circle is $x^2+y^2-8x-12y+\alpha=0$. The center is $(4, 6)$ and the radius $r = \sqrt{4^2+6^2-\alpha} = \sqrt{52-\alpha}$.
Since the circle lies in the first quadrant without touching the axes,the distance from the center to the axes must be greater than the radius: $r < 4$ and $r < 6$. Thus,$r < 4$,which implies $\sqrt{52-\alpha} < 4$ $\Rightarrow 52-\alpha < 16$ $\Rightarrow \alpha > 36$.
Since $(6, 6)$ is an interior point,substituting it into the circle equation gives $6^2+6^2-8(6)-12(6)+\alpha < 0$.
$36+36-48-72+\alpha < 0$ $\Rightarrow \alpha - 48 < 0$ $\Rightarrow \alpha < 48$.
Combining these,we get $36 < \alpha < 48$.

(B) The power of a point $P(x_1, y_1)$ with respect to a circle $(x - h)^2 + (y - k)^2 = r^2$ is given by the formula $P = (x_1 - h)^2 + (y_1 - k)^2 - r^2$.
Given the point $P(-3, 7)$,the centre $(h, k) = (3, 7)$,and the radius $r = 2$.
Substituting these values into the formula:
$P = (-3 - 3)^2 + (7 - 7)^2 - 2^2$
$P = (-6)^2 + (0)^2 - 4$
$P = 36 + 0 - 4$
$P = 32$.
Thus,the power of the point is $32$.

(D) Let $(h, k)$ be the center of the circle.
Since the circle touches the $y$-axis at $(0,2)$,the radius $r = |h|$ and the $y$-coordinate of the center $k = 2$.
The equation of the circle is $(x-h)^2 + (y-2)^2 = h^2$.
Since it passes through $(-1, 0)$,we substitute these coordinates:
$(-1-h)^2 + (0-2)^2 = h^2$
$1 + 2h + h^2 + 4 = h^2$
$2h = -5 \Rightarrow h = -\frac{5}{2}$.
The equation of the circle is $\left(x + \frac{5}{2}\right)^2 + (y-2)^2 = \left(\frac{5}{2}\right)^2$.
Checking option $(d)$ $(-4, 0)$:
$\left(-4 + \frac{5}{2}\right)^2 + (0-2)^2 = \left(-\frac{3}{2}\right)^2 + 4 = \frac{9}{4} + 4 = \frac{25}{4} = \left(\frac{5}{2}\right)^2$.
Thus,the circle passes through $(-4, 0)$.

(C) The given circle is $3x^2+3y^2+x+y-1=0$.
Dividing by $3$,we get $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$.
The center is $(h, k) = \left(-\frac{1}{6}, -\frac{1}{6}\right)$ and the radius $r$ is given by $r^2 = h^2+k^2-c = \frac{1}{36}+\frac{1}{36}+\frac{1}{3} = \frac{1+1+12}{36} = \frac{14}{36} = \frac{7}{18}$.
The length of the tangent $L$ from point $(2,-2)$ to the circle $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$ is given by $L^2 = x_1^2+y_1^2+\frac{1}{3}x_1+\frac{1}{3}y_1-\frac{1}{3}$.
$L^2 = (2)^2+(-2)^2+\frac{1}{3}(2)+\frac{1}{3}(-2)-\frac{1}{3} = 4+4+\frac{2}{3}-\frac{2}{3}-\frac{1}{3} = 8-\frac{1}{3} = \frac{23}{3}$.
Since the radius of circle $S$ is $L$,the radius squared is $R^2 = \frac{23}{3}$.
The equation of circle $S$ is $(x+\frac{1}{6})^2+(y+\frac{1}{6})^2 = \frac{23}{3}$.
The power of the point $(2,1)$ with respect to circle $S$ is $(x_1+\frac{1}{6})^2+(y_1+\frac{1}{6})^2 - R^2$.
$= (2+\frac{1}{6})^2+(1+\frac{1}{6})^2 - \frac{23}{3} = (\frac{13}{6})^2+(\frac{7}{6})^2 - \frac{23}{3} = \frac{169}{36}+\frac{49}{36} - \frac{276}{36} = \frac{218-276}{36} = -\frac{58}{36} = -\frac{29}{18}$.

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,where the radius $r = \sqrt{g^2+f^2-c}$.
For the first circle $C_1: x^2+y^2-6x+12y+15=0$,we have $g=-3, f=6, c=15$.
$r_1 = \sqrt{(-3)^2+6^2-15} = \sqrt{9+36-15} = \sqrt{30}$.
For the second circle $C_2: x^2+y^2-6x+12y-15=0$,we have $g=-3, f=6, c=-15$.
$r_2 = \sqrt{(-3)^2+6^2-(-15)} = \sqrt{9+36+15} = \sqrt{60}$.
The ratio of the areas is $\frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{30}{60} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(C) Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(c, c)$ and its radius is $c$,where $c > 0$.
The equation of the circle is $(x-c)^2 + (y-c)^2 = c^2$.
The line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.
Since the circle touches this line,the perpendicular distance from the center $(c, c)$ to the line must be equal to the radius $c$.
$\frac{|4c + 3c - 12|}{\sqrt{4^2 + 3^2}} = c$
$\frac{|7c - 12|}{5} = c$
$|7c - 12| = 5c$
Case $1$: $7c - 12 = 5c \implies 2c = 12 \implies c = 6$.
Case $2$: $7c - 12 = -5c \implies 12c = 12 \implies c = 1$.
Thus,$c = 1$ or $c = 6$.

(D) The power of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Given the point $(1, 6)$ and the circle $x^2 + y^2 + 4x - 6y - a = 0$,the power is $-16$.
Substituting the values:
$(1)^2 + (6)^2 + 4(1) - 6(6) - a = -16$
$1 + 36 + 4 - 36 - a = -16$
$5 - a = -16$
$-a = -16 - 5$
$-a = -21$
$a = 21$
Thus,the correct option is $D$.

(C) The center $(h, k)$ of the circle is the intersection of the diameters $x+2y-2=0$ and $2x+3y-1=0$.
Solving these equations:
$x+2y=2$ $(1)$
$2x+3y=1$ $(2)$
Multiplying $(1)$ by $2$: $2x+4y=4$ $(3)$
Subtracting $(2)$ from $(3)$: $(2x+4y)-(2x+3y) = 4-1$,so $y=3$.
Substituting $y=3$ in $(1)$: $x+2(3)=2 \implies x+6=2 \implies x=-4$.
So,the center is $(-4, 3)$.
The radius $R$ is the distance between the center $(-4, 3)$ and the point $(8, 8)$:
$R^2 = (8 - (-4))^2 + (8 - 3)^2 = 12^2 + 5^2 = 144 + 25 = 169$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = R^2$:
$(x + 4)^2 + (y - 3)^2 = 169$
$x^2 + 8x + 16 + y^2 - 6y + 9 = 169$
$x^2 + y^2 + 8x - 6y - 144 = 0$.
Comparing with $x^2 + y^2 + px + qy + r = 0$,we get $p=8, q=-6, r=-144$.
Then $p^2 + q^2 + r = 8^2 + (-6)^2 - 144 = 64 + 36 - 144 = 100 - 144 = -44$.

(A) The given equation $|x-2|+|y-3|=4$ represents a square with vertices at $(2+4, 3) = (6, 3)$,$(2-4, 3) = (-2, 3)$,$(2, 3+4) = (2, 7)$,and $(2, 3-4) = (2, -1)$.
This square is centered at $(2, 3)$ and has a side length of $4\sqrt{2}$.
The circle touching these lines is the incircle of this square.
The center of the circle is the same as the center of the square,which is $(h, k) = (2, 3)$.
The radius $r$ of the incircle is half the distance between parallel sides,or the perpendicular distance from the center $(2, 3)$ to any of the lines,such as $(x-2)+(y-3)=4 \implies x+y-9=0$.
The distance $r = \frac{|2+3-9|}{\sqrt{1^2+1^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$.
The equation of the circle is $(x-h)^2+(y-k)^2=r^2$.
Substituting the values: $(x-2)^2+(y-3)^2=(2\sqrt{2})^2$.
$x^2-4x+4+y^2-6y+9=8$.
$x^2+y^2-4x-6y+5=0$.

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through $(2,0)$,we have $4+4g+c=0$,so $c = -4-4g$.
The intercept on the $X$-axis is given by $2\sqrt{g^2-c} = 5$.
Squaring both sides,$4(g^2-c) = 25$.
Substituting $c = -4-4g$,we get $4(g^2+4g+4) = 25$,which simplifies to $4g^2+16g+16=25$,or $4g^2+16g-9=0$.
Solving for $g$,$g = \frac{-16 \pm \sqrt{256-4(4)(-9)}}{8} = \frac{-16 \pm \sqrt{256+144}}{8} = \frac{-16 \pm 20}{8}$.
Thus,$g = \frac{4}{8} = 0.5$ or $g = \frac{-36}{8} = -4.5$.
Since the centre $(-g, -f)$ lies in the first quadrant,$-g > 0$,so $g < 0$. Thus $g = -4.5 = -9/2$.
Then $c = -4-4(-4.5) = -4+18 = 14$.
The equation is $x^2+y^2-9x+2fy+14=0$.
Let $2f = -2k$,where $k \in R^{+}$,so the equation is $x^2+y^2-9x-2ky+14=0$.

(C) In $\triangle ABC$,since $\angle A = 90^{\circ}$,the side $BC$ acts as the diameter of the circumcircle.
Given the endpoints of the diameter are $B(2, -4)$ and $C(1, 5)$,the equation of the circle is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the coordinates:
$(x - 2)(x - 1) + (y - (-4))(y - 5) = 0$
$(x^2 - x - 2x + 2) + (y^2 - 5y + 4y - 20) = 0$
$x^2 + y^2 - 3x - y - 18 = 0$.
Thus,option $C$ is correct.

(C) Given,$\left|\frac{z-3}{z-2i}\right|=2$.
Substituting $z=x+iy$,we get $|(x-3)+iy| = 2|x+i(y-2)|$.
Squaring both sides,we have $(x-3)^2 + y^2 = 4(x^2 + (y-2)^2)$.
Expanding the terms,$x^2 - 6x + 9 + y^2 = 4(x^2 + y^2 - 4y + 4)$.
$x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 - 16y + 16$.
Rearranging the terms,$3x^2 + 3y^2 + 6x - 16y + 7 = 0$.
Dividing by $3$,we get $x^2 + y^2 + 2x - \frac{16}{3}y + \frac{7}{3} = 0$.
Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = 2 \Rightarrow g = 1$ and $2f = -\frac{16}{3} \Rightarrow f = -\frac{8}{3}$.
The centre of the circle is $(-g, -f) = (-1, \frac{8}{3})$.

(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(0,0)$,we have $0 + 0 + 0 + 0 + c = 0$,so $c = 0$.
Since it passes through $(2,0)$,we have $4 + 0 + 4g + 0 + 0 = 0$,which gives $g = -1$.
Since it passes through $(0,-2)$,we have $0 + 4 + 0 - 4f + 0 = 0$,which gives $f = 1$.
The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$.
For the point $(k,-2)$ to lie on this circle,we substitute the coordinates into the equation:
$k^2 + (-2)^2 - 2(k) + 2(-2) = 0$
$k^2 + 4 - 2k - 4 = 0$
$k^2 - 2k = 0$
$k(k - 2) = 0$
Since the points must be distinct,$k$ cannot be $0$ (as $(0,-2)$ is already a point).
Therefore,$k = 2$.

(B) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,with radius $r = \sqrt{g^2+f^2-c}$.
For the first circle $x^2+y^2+8x-6y=0$,$g=4, f=-3, c=0$. Thus,$r_1 = \sqrt{4^2+(-3)^2-0} = \sqrt{16+9} = 5$. Perimeter $P_1 = 2\pi r_1 = 10\pi$.
For the second circle $4x^2+4y^2-4x-12y-186=0$,divide by $4$: $x^2+y^2-x-3y-46.5=0$. Here $g=-0.5, f=-1.5, c=-46.5$. Thus,$r_2 = \sqrt{(-0.5)^2+(-1.5)^2-(-46.5)} = \sqrt{0.25+2.25+46.5} = \sqrt{49} = 7$. Perimeter $P_2 = 2\pi r_2 = 14\pi$.
For the third circle $x^2+y^2-6x+6y-9=0$,$g=-3, f=3, c=-9$. Thus,$r_3 = \sqrt{(-3)^2+3^2-(-9)} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \approx 3 \times 1.732 = 5.196$. Perimeter $P_3 = 2\pi r_3 = 6\sqrt{3}\pi \approx 10.39\pi$.
Comparing the perimeters: $10\pi < 10.39\pi < 14\pi$,which implies $P_1 < P_3 < P_2$.

(C) Let the equation of the circle be $x^2+y^2+2gx'+2fy'+c=0$.
Since the circle passes through $(-1,0)$,$(-1,1)$,and $(1,1)$,we substitute these points into the equation:
$1$) For $(-1,0)$: $(-1)^2+0^2+2g(-1)+2f(0)+c=0 \implies 1-2g+c=0 \implies c=2g-1$.
$2$) For $(-1,1)$: $(-1)^2+1^2+2g(-1)+2f(1)+c=0 \implies 2-2g+2f+c=0$.
Substituting $c=2g-1$ into this: $2-2g+2f+2g-1=0 \implies 2f+1=0 \implies f=-1/2$.
$3$) For $(1,1)$: $1^2+1^2+2g(1)+2f(1)+c=0 \implies 2+2g+2f+c=0$.
Substituting $f=-1/2$ and $c=2g-1$: $2+2g+2(-1/2)+2g-1=0 \implies 2+2g-1+2g-1=0 \implies 4g=0 \implies g=0$.
Then $c=2(0)-1=-1$.
The equation is $x^2+y^2+0x+2(-1/2)y-1=0 \implies x^2+y^2-y-1=0$.
To match the form $ax^2+ay^2+2gx+2fy-2=0$,we multiply the entire equation by $2$:
$2x^2+2y^2-2y-2=0$.
Comparing $2x^2+2y^2+0x-2y-2=0$ with $ax^2+ay^2+2gx+2fy-2=0$,we get $a=2$.

(A) Let the points be $A(1, 1), B(-2, 2)$,and $C(2, -2)$.
Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting the points into the equation:
For $(1, 1): 1 + 1 + 2g + 2f + c = 0 \Rightarrow 2g + 2f + c = -2$ $(i)$
For $(-2, 2): 4 + 4 - 4g + 4f + c = 0 \Rightarrow -4g + 4f + c = -8$ $(ii)$
For $(2, -2): 4 + 4 + 4g - 4f + c = 0 \Rightarrow 4g - 4f + c = -8$ $(iii)$
Subtracting $(iii)$ from $(ii): -8g + 8f = 0 \Rightarrow g = f$.
Substituting $g = f$ in $(i)$ and $(ii): 4g + c = -2$ and $-4g + 4g + c = -8 \Rightarrow c = -8$.
Substituting $c = -8$ in $4g + c = -2: 4g - 8 = -2$ $\Rightarrow 4g = 6$ $\Rightarrow g = \frac{3}{2}$.
Thus,$g = f = \frac{3}{2}$ and $c = -8$.
The centre of the circle is $(-g, -f) = \left(-\frac{3}{2}, -\frac{3}{2}\right)$.
The perpendicular distance from $\left(-\frac{3}{2}, -\frac{3}{2}\right)$ to the line $3x - 4y + 1 = 0$ is given by $d = \frac{|3(-\frac{3}{2}) - 4(-\frac{3}{2}) + 1|}{\sqrt{3^2 + (-4)^2}}$.
$d = \frac{|-\frac{9}{2} + 6 + 1|}{5} = \frac{|\frac{5}{2}|}{5} = \frac{1}{2}$.

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the points $(-1, 1), (0, -1)$,and $(1, 0)$ lie on the circle:
For $(-1, 1): 1 + 1 - 2g + 2f + c = 0 \Rightarrow -2g + 2f + c = -2 \dots (i)$
For $(0, -1): 0 + 1 + 0 - 2f + c = 0 \Rightarrow -2f + c = -1 \dots (ii)$
For $(1, 0): 1 + 0 + 2g + 0 + c = 0 \Rightarrow 2g + c = -1 \dots (iii)$
Adding $(ii)$ and $(iii)$,we get $2c = -2 \Rightarrow c = -1$.
Substituting $c = -1$ into $(iii)$,$2g - 1 = -1 \Rightarrow g = 0$.
Substituting $c = -1$ into $(ii)$,$-2f - 1 = -1 \Rightarrow f = 0$.
The equation of the circle is $x^2 + y^2 - 1 = 0$.
Since $(1, k)$ lies on the circle:
$1^2 + k^2 - 1 = 0
$ $\Rightarrow k^2 = 0
$ $\Rightarrow k = 0$.
However,the problem states $k \neq 0$. Re-evaluating the points:
For $(-1, 1): 1 + 1 - 2g + 2f + c = 0 \Rightarrow -2g + 2f + c = -2$.
For $(0, -1): 1 - 2f + c = 0 \Rightarrow -2f + c = -1$.
For $(1, 0): 1 + 2g + c = 0 \Rightarrow 2g + c = -1$.
Adding $(ii)$ and $(iii)$: $2c = -2 \Rightarrow c = -1$.
$2g = -1 - (-1) = 0 \Rightarrow g = 0$.
$-2f = -1 - (-1) = 0 \Rightarrow f = 0$.
The circle is $x^2 + y^2 = 1$. The point $(1, k)$ on this circle implies $1 + k^2 = 1 \Rightarrow k = 0$.
Given the options,there is a discrepancy in the provided points or the question statement. Assuming the intended circle equation is $x^2 + y^2 - x - y - 1 = 0$ or similar,but based on the provided points,$k=0$ is the only solution.

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the points $(2,0)$ and $(1,2)$ lie on the circle:
$4 + 4g + c = 0 \Rightarrow 4g + c = -4$ $(i)$
$1 + 4 + 2g + 4f + c = 0 \Rightarrow 2g + 4f + c = -5$ $(ii)$
The power of the point $(0,2)$ with respect to the circle is given by $S(0,2) = 0^2 + 2^2 + 2g(0) + 2f(2) + c = 4 + 4f + c$.
Given the power is $4$,we have $4 + 4f + c = 4 \Rightarrow 4f + c = 0$ $(iii)$
Subtracting $(iii)$ from $(ii)$: $(2g + 4f + c) - (4f + c) = -5 - 0$ $\Rightarrow 2g = -5$ $\Rightarrow g = -\frac{5}{2}$.
Substituting $g$ into $(i)$: $4(-\frac{5}{2}) + c = -4$ $\Rightarrow -10 + c = -4$ $\Rightarrow c = 6$.
Substituting $c$ into $(iii)$: $4f + 6 = 0 \Rightarrow f = -\frac{6}{4} = -\frac{3}{2}$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{5}{2})^2 + (-\frac{3}{2})^2 - 6} = \sqrt{\frac{25}{4} + \frac{9}{4} - 6} = \sqrt{\frac{34}{4} - 6} = \sqrt{\frac{17}{2} - 6} = \sqrt{\frac{5}{2}}$.

(D) The center of the circle is the intersection point of its diameters $x - y = 2$ and $2x + 3y = 14$.
Solving $x - y = 2$ for $y$,we get $y = x - 2$.
Substituting this into the second equation: $2x + 3(x - 2) = 14$.
$2x + 3x - 6 = 14$ $\Rightarrow 5x = 20$ $\Rightarrow x = 4$.
Then $y = 4 - 2 = 2$.
So,the center of the circle is $(4, 2)$.
The circle passes through $(1, -2)$. The radius $r$ is the distance between the center $(4, 2)$ and the point $(1, -2)$.
$r = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(A) Let the centre of the circle be $C(\alpha, \beta)$. Since the centre lies on the line $x-y+1=0$,we have $\beta = \alpha+1$. Thus,$C = (\alpha, \alpha+1)$.
Given points $P(3,1)$ and $Q(-2,4)$ lie on the circle,so $CP^2 = CQ^2$.
$(\alpha-3)^2 + (\alpha+1-1)^2 = (\alpha+2)^2 + (\alpha+1-4)^2$
$(\alpha-3)^2 + \alpha^2 = (\alpha+2)^2 + (\alpha-3)^2$
$\alpha^2 = (\alpha+2)^2$
$\alpha^2 = \alpha^2 + 4\alpha + 4$
$4\alpha = -4 \Rightarrow \alpha = -1$.
So,the centre is $C(-1, 0)$.
The radius $r$ is the distance $CP = \sqrt{(-1-3)^2 + (0-1)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16+1} = \sqrt{17}$.
The parametric equations of a circle with centre $(h, k)$ and radius $r$ are $x = h + r \cos \theta$ and $y = k + r \sin \theta$.
Substituting $h=-1, k=0, r=\sqrt{17}$,we get $x = -1 + \sqrt{17} \cos \theta$ and $y = \sqrt{17} \sin \theta$.

(B) The circle passes through the points $A(3,4)$,$B(3,2)$,and $C(1,4)$.
Since $AB$ is a vertical line segment $(x=3)$ and $AC$ is a horizontal line segment $(y=4)$,the angle at $A(3,4)$ is $90^\circ$.
Thus,$BC$ is the diameter of the circle.
The midpoint of $BC$ is the center $(h, k) = (\frac{3+1}{2}, \frac{2+4}{2}) = (2, 3)$.
The radius $r$ is the distance from the center $(2, 3)$ to $(3, 4)$:
$r = \sqrt{(3-2)^2 + (4-3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The parametric equations are $x = 2 + \sqrt{2} \cos \theta$ and $y = 3 + \sqrt{2} \sin \theta$.
Comparing with $x = a + r \cos \theta$ and $y = b + r \sin \theta$,we get $a = 2$,$b = 3$,and $r = \sqrt{2}$.
Then,$b^a \cdot r^a = 3^2 \cdot (\sqrt{2})^2 = 9 \cdot 2 = 18$.