System of circles Questions in English

(D) The given equation of the coaxial system is $x^2+y^2+2 \lambda x+c=0$.
Limiting points are the centers of the point circles in the system.
$A$ point circle is obtained when the radius $r=0$.
The radius $r$ of the circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{g^2+f^2-c}$.
Here,$g=\lambda$,$f=0$,and the constant term is $c$.
Thus,$r = \sqrt{\lambda^2-c}$.
For the limiting points to be distinct,the radius must be imaginary,which implies $\lambda^2-c < 0$,or $c > \lambda^2$.
However,for the system to have limiting points that are real and distinct,the condition is $c > 0$.

(C) The equations of the circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$.
The family of circles is given by $S_1 + \lambda(S_1 - S_2) = 0$ or $S_1 + k S_2 = 0$.
The limiting points are the centers of the point circles in the co-axial system.
$A$ circle $x^2+y^2+2gx+2fy+c=0$ is a point circle if $g^2+f^2-c=0$.
The radical axis is $S_1 - S_2 = 0$ $\Rightarrow (2 + \frac{2}{5})x + (-2 + \frac{16}{5})y + (2 - \frac{13}{5}) = 0$ $\Rightarrow \frac{12}{5}x + \frac{6}{5}y - \frac{3}{5} = 0$ $\Rightarrow 4x + 2y - 1 = 0$.
Any circle in the system is $S_1 + \lambda(4x+2y-1) = 0 \Rightarrow x^2+y^2+(2+4\lambda)x + (-2+2\lambda)y + (2-\lambda) = 0$.
For a point circle,$g^2+f^2-c=0 \Rightarrow (1+2\lambda)^2 + (-1+\lambda)^2 - (2-\lambda) = 0$.
$1+4\lambda+4\lambda^2 + 1-2\lambda+\lambda^2 - 2+\lambda = 0 \Rightarrow 5\lambda^2+3\lambda = 0$.
Thus,$\lambda = 0$ or $\lambda = -\frac{3}{5}$.
For $\lambda = 0$,the center is $(-g, -f) = (-1, 1)$.
For $\lambda = -\frac{3}{5}$,the center is $(-(1+2(-\frac{3}{5})), -(-1+(-\frac{3}{5}))) = (- (1-\frac{6}{5}), -(-\frac{8}{5})) = (\frac{1}{5}, \frac{8}{5})$.
The limiting points are $(-1, 1)$ and $(\frac{1}{5}, \frac{8}{5})$.

(C) The equation of a circle passing through the origin $(0,0)$ and cutting the circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ orthogonally is given by the determinant equation:
$\left|\begin{array}{ccc} x^2+y^2 & x & y \\ c_1 & g_1 & f_1 \\ c_2 & g_2 & f_2 \end{array}\right| = 0$
Here,$c_1=12, g_1=2, f_1=3$ and $c_2=9, g_2=2, f_2=-3$.
Substituting these values:
$\left|\begin{array}{ccc} x^2+y^2 & x & y \\ 12 & 2 & 3 \\ 9 & 2 & -3 \end{array}\right| = 0$
Expanding the determinant:
$(x^2+y^2)(-6-6) - x(-36-27) + y(24-18) = 0$
$-12(x^2+y^2) + 63x + 6y = 0$
$x^2+y^2 - \frac{63}{12}x - \frac{6}{12}y = 0$
$x^2+y^2 - \frac{21}{4}x - \frac{1}{2}y = 0$
The centre $(h, k)$ is $(\frac{21}{8}, \frac{1}{4})$.
Thus,$k-2h = \frac{1}{4} - 2(\frac{21}{8}) = \frac{1}{4} - \frac{21}{4} = -\frac{20}{4} = -5$.

(B) Let the equation of circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts $C_1$ and $C_2$ orthogonally,we have:
$2g(-4) + 2f(-1) = c + 16 \implies -8g - 2f = c + 16 \quad (i)$
$2g(-2) + 2f(-2) = c - 1 \implies -4g - 4f = c - 1 \quad (ii)$
Subtracting $(ii)$ from $(i)$: $-4g + 2f = 17 \implies 2f = 4g + 17 \implies f = 2g + 8.5$.
The common chord of $S=0$ and $C_1=0$ is $S - C_1 = 0$:
$(2g+8)x + (2f+2)y + (c-16) = 0$.
Comparing this with the given chord $2x + 13y - 15 = 0$:
$\frac{2g+8}{2} = \frac{2f+2}{13} = \frac{c-16}{-15} = k$.
$2g+8 = 2k \implies g = k-4$.
$2f+2 = 13k \implies f = \frac{13k-2}{2}$.
Substitute into $2f = 4g + 17$: $13k-2 = 4(k-4) + 17 \implies 13k-2 = 4k-16+17 \implies 9k = 3 \implies k = \frac{1}{3}$.
Then $g = \frac{1}{3} - 4 = \frac{-11}{3}$ and $f = \frac{13(1/3)-2}{2} = \frac{7/3}{2} = \frac{7}{6}$.
The centre of $S$ is $(-g, -f) = \left(\frac{11}{3}, \frac{-7}{6}\right)$.

(A) The given circle is $S: x^2+y^2-4x-6y-12=0$. The center $C_1$ is $(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$.
Let the required circle have center $C_2(h, k)$ and radius $r_2 = 3$.
Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 5 : 3$.
Using the section formula for external division:
$-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$.
$-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$.
The equation of the circle is $(x - \frac{4}{5})^2 + (y - \frac{7}{5})^2 = 3^2$.
$x^2 - \frac{8}{5}x + \frac{16}{25} + y^2 - \frac{14}{5}y + \frac{49}{25} = 9$.
$x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{65}{25} - 9 = 0$.
$x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y + \frac{13}{5} - 9 = 0$.
$x^2 + y^2 - \frac{8}{5}x - \frac{14}{5}y - \frac{32}{5} = 0$.
Multiplying by $5$,we get $5x^2 + 5y^2 - 8x - 14y - 32 = 0$.

(A) The circle $S$ is $x^2+y^2-14x+6y+33=0$.
To find the intersection with the $X$-axis,set $y=0$: $x^2-14x+33=0 \implies (x-3)(x-11)=0$.
Thus,the points are $A(3, 0)$ and $B(11, 0)$ since $OB > OA$.
The midpoint $C$ of $AB$ is $(\frac{3+11}{2}, 0) = (7, 0)$.
The line $L$ passes through $(7, 0)$ with slope $-1$: $y-0 = -1(x-7) \implies x+y-7=0$.
Since $L$ is the radical axis of $S$ and $S^{\prime}$,the equation of $S^{\prime}$ is $S + kL = 0$:
$x^2+y^2-14x+6y+33 + k(x+y-7) = 0$.
$x^2+y^2+(k-14)x+(k+6)y+(33-7k) = 0$.
The center of $S^{\prime}$ is $(-\frac{k-14}{2}, -\frac{k+6}{2})$.
Since $L$ is the diameter of $S^{\prime}$,the center must lie on $L$:
$-\frac{k-14}{2} - \frac{k+6}{2} - 7 = 0 \implies -k+14-k-6-14 = 0 \implies -2k-6=0 \implies k=-3$.
Substituting $k=-3$ into the equation:
$x^2+y^2+(-3-14)x+(-3+6)y+(33-7(-3)) = 0$.
$x^2+y^2-17x+3y+54=0$.

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ which is given as $(\alpha, \beta)$,so $g=-\alpha$ and $f=-\beta$. The circle $x^2+y^2+2gx+2fy+c=0$ cuts $x^2+y^2-2y-3=0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and $g_2=0, f_2=-1, c_2=-3$. So,$2g(0) + 2f(-1) = c-3 \implies -2f = c-3 \implies c = 3-2f = 3+2\beta$.
Next,it cuts $x^2+y^2+4x+3=0$ orthogonally. Here $g_3=2, f_3=0, c_3=3$. So,$2g(2) + 2f(0) = c+3 \implies 4g = c+3 \implies 4(-\alpha) = c+3 \implies c = -4\alpha-3$.
Equating the two expressions for $c$: $3+2\beta = -4\alpha-3 \implies 4\alpha+2\beta = -6 \implies 2\alpha+\beta = -3$.
Since the centre $(\alpha, \beta)$ lies on $2x-3y+4=0$,we have $2\alpha-3\beta+4=0$.
From $2\alpha+\beta = -3$,we get $2\alpha = -3-\beta$. Substituting this into the line equation: $(-3-\beta)-3\beta+4=0 \implies -4\beta+1=0 \implies \beta=1/4$.
Then $2\alpha = -3-1/4 = -13/4 \implies \alpha = -13/8$.
The value $2\alpha+\beta = -3$.

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ where $\alpha = -g$ and $\beta = -f$.
Since the circle cuts $x^2+y^2+2x-3y-5=0$ orthogonally,$2g(1) + 2f(-3/2) = c-5 \implies 2g-3f = c-5$.
Since the circle cuts $x^2+y^2-3x+2y+1=0$ orthogonally,$2g(-3/2) + 2f(1) = c+1 \implies -3g+2f = c+1$.
Subtracting the two equations: $(2g-3f) - (-3g+2f) = (c-5) - (c+1) \implies 5g-5f = -6 \implies g-f = -6/5$.
Since the circle passes through $(1, -1)$,$1+1+2g-2f+c=0 \implies 2g-2f+c = -2$.
From $2g-3f = c-5$,we have $c = 2g-3f+5$. Substituting into the circle equation: $2g-2f+(2g-3f+5) = -2 \implies 4g-5f = -7$.
We have the system: $g-f = -1.2$ and $4g-5f = -7$. Multiplying the first by $4$: $4g-4f = -4.8$.
Subtracting: $(4g-5f) - (4g-4f) = -7 - (-4.8) \implies -f = -2.2 \implies f = 2.2$.
Then $g = 2.2 - 1.2 = 1$.
Thus,$\alpha = -g = -1$ and $\beta = -f = -2.2$.
Therefore,$\alpha-5\beta = -1 - 5(-2.2) = -1 + 11 = 10$.

(C) The equation of the family of circles passing through the intersection of the circle $S_1: x^2+y^2+2x-4y+4=0$ and the line $L: x+y-2=0$ is given by $S_1 + \lambda L = 0$.
$x^2+y^2+2x-4y+4 + \lambda(x+y-2) = 0$
$x^2+y^2+(2+\lambda)x + (\lambda-4)y + (4-2\lambda) = 0$.
This circle is orthogonal to $S_2: x^2+y^2-2x-4y-4=0$.
The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = \frac{2+\lambda}{2}$,$f_1 = \frac{\lambda-4}{2}$,$c_1 = 4-2\lambda$ and $g_2 = -1$,$f_2 = -2$,$c_2 = -4$.
Substituting these values: $2(\frac{2+\lambda}{2})(-1) + 2(\frac{\lambda-4}{2})(-2) = 4-2\lambda - 4$.
$-(2+\lambda) - 2(\lambda-4) = -2\lambda$.
$-2-\lambda - 2\lambda + 8 = -2\lambda$.
$6 - 3\lambda = -2\lambda \implies \lambda = 6$.
Substituting $\lambda = 6$ into the circle equation: $x^2+y^2+8x+2y-8=0$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{4^2+1^2-(-8)} = \sqrt{16+1+8} = \sqrt{25} = 5$.

(A) The given circle is $x^2+y^2-6x-12y+1=0$. Comparing this with $x^2+y^2+2g_1x+2f_1y+c_1=0$,we get $g_1=-3, f_1=-6, c_1=1$.
Let the circle $C$ be $x^2+y^2+2g_2x+2f_2y+c_2=0$. Given the centre is $(-4, 2)$,so $-g_2=-4 \Rightarrow g_2=4$ and $-f_2=2 \Rightarrow f_2=-2$.
Two circles cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
Substituting the values: $2(-3)(4) + 2(-6)(-2) = 1 + c_2$.
$-24 + 24 = 1 + c_2 \Rightarrow c_2 = -1$.
The radius $r$ of circle $C$ is given by $\sqrt{g_2^2+f_2^2-c_2}$.
$r = \sqrt{4^2+(-2)^2-(-1)} = \sqrt{16+4+1} = \sqrt{21}$.

(B) Let the required circle be $x^2+y^2+2gx+2fy+c=0$.
Since it cuts the given circles orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ must be satisfied for each.
For $x^2+y^2-4x-4y+3=0$: $2g(-2) + 2f(-2) = c + 3 \implies -4g - 4f = c + 3$.
For $x^2+y^2+4x-4y+3=0$: $2g(2) + 2f(-2) = c + 3 \implies 4g - 4f = c + 3$.
For $x^2+y^2+4x+4y+3=0$: $2g(2) + 2f(2) = c + 3 \implies 4g + 4f = c + 3$.
Subtracting the first two equations: $8g = 0 \implies g = 0$.
Subtracting the last two equations: $8f = 0 \implies f = 0$.
Substituting $g=0$ and $f=0$ into any equation: $0 = c + 3 \implies c = -3$.
The equation of the circle is $x^2+y^2-3=0$,which is $x^2+y^2=3$.
The radius is $\sqrt{r^2} = \sqrt{3}$.

(D) The condition that a circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts another circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$ at the extremities of its diameter is equivalent to the condition of orthogonality: $2(gg_{i}+ff_{i})=c+c_{i}$.
For the first circle $x^{2}+y^{2}-5=0$,$g_{1}=0, f_{1}=0, c_{1}=-5$. Applying the condition: $2(g(0)+f(0))=c-5 \Rightarrow c=5$.
For the second circle $x^{2}+y^{2}-8x-6y+10=0$,$g_{2}=-4, f_{2}=-3, c_{2}=10$. Applying the condition: $2(g(-4)+f(-3))=c+10$ $\Rightarrow 2(-4g-3f)=5+10$ $\Rightarrow 4g+3f=-15/2$.
For the third circle $x^{2}+y^{2}-4x+2y-2=0$,$g_{3}=-2, f_{3}=1, c_{3}=-2$. Applying the condition: $2(g(-2)+f(1))=c-2$ $\Rightarrow 2(-2g+f)=5-2$ $\Rightarrow -2g+f=3/2$.
Solving the system of equations $4g+3f=-7.5$ and $-2g+f=1.5$:
From the second equation,$f=2g+1.5$. Substituting into the first: $4g+3(2g+1.5)=-7.5$ $\Rightarrow 10g+4.5=-7.5$ $\Rightarrow 10g=-12$ $\Rightarrow g=-1.2$.
Then $f=2(-1.2)+1.5 = -2.4+1.5 = -0.9$.
Checking option $D$: $4f = 4(-0.9) = -3.6$ and $3g = 3(-1.2) = -3.6$. Thus,$4f=3g$ is correct.

(A) Given circle equation: $x^{2}+y^{2}-4x-6y+9=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-2, f=-3, c=9$.
Centre $B = (-g, -f) = (2, 3)$.
Radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-2)^{2}+(-3)^{2}-9} = \sqrt{4+9-9} = 2$.
Let the centre of the required circle be $A(1, 1)$.
The distance $AB$ between the centres is $AB = \sqrt{(2-1)^{2}+(3-1)^{2}} = \sqrt{1^{2}+2^{2}} = \sqrt{5}$.
Since the diameter of the first circle is a chord of the second circle,the radius $R$ of the second circle is the hypotenuse of the right-angled triangle formed by the distance between centres and the radius of the first circle.
$R = \sqrt{AB^{2}+r^{2}} = \sqrt{(\sqrt{5})^{2}+2^{2}} = \sqrt{5+4} = \sqrt{9} = 3$.

(B) Let the equation of the circles be $x^2+y^2+2gx+2fy+d=0 \quad \ldots(1)$.
Since these circles pass through $(0, a)$ and $(0, -a)$,we have $a^2+2fa+d=0 \quad \ldots(2)$ and $a^2-2fa+d=0 \quad \ldots(3)$.
Solving $(2)$ and $(3)$,we get $f=0$ and $d=-a^2$.
Substituting these values in $(1)$,we obtain $x^2+y^2+2gx-a^2=0 \quad \ldots(4)$.
Since the line $y=mx+c$ touches this circle,the perpendicular distance from the centre $(-g, 0)$ to the line $mx-y+c=0$ is equal to the radius $\sqrt{g^2+a^2}$.
Thus,$\frac{|-mg+c|}{\sqrt{1+m^2}} = \sqrt{g^2+a^2}$.
Squaring both sides,we get $(c-mg)^2 = (1+m^2)(g^2+a^2)$.
Expanding this,$c^2-2mcg+m^2g^2 = g^2+a^2+m^2g^2+m^2a^2$.
Rearranging,$g^2 + 2mcg + a^2(1+m^2) - c^2 = 0$.
Let $g_1$ and $g_2$ be the roots of this quadratic equation in $g$.
Then the product of the roots is $g_1g_2 = a^2(1+m^2)-c^2 \quad \ldots(5)$.
The two circles are $x^2+y^2+2g_1x-a^2=0$ and $x^2+y^2+2g_2x-a^2=0$.
For these circles to cut orthogonally,$2g_1g_2 + 2f_1f_2 = d_1+d_2$.
Here $f_1=f_2=0$ and $d_1=d_2=-a^2$,so $2g_1g_2 = -2a^2$,which implies $g_1g_2 = -a^2 \quad \ldots(6)$.
Comparing $(5)$ and $(6)$,$-a^2 = a^2(1+m^2) - c^2$.
Therefore,$c^2 = a^2(1+m^2) + a^2 = a^2(2+m^2)$.

(A) Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ intersect orthogonally if and only if $2(g_{1}g_{2}+f_{1}f_{2})=c_{1}+c_{2}$.
For the given circles:
Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$
Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$
Substituting these values into the condition:
$2((1)(0) + (k)(k)) = 6 + k$
$2k^{2} = 6 + k$
$2k^{2} - k - 6 = 0$
Factoring the quadratic equation:
$2k^{2} - 4k + 3k - 6 = 0$
$2k(k-2) + 3(k-2) = 0$
$(k-2)(2k+3) = 0$
Thus,$k = 2$ or $k = -\frac{3}{2}$.

(B) For two circles to intersect at two distinct points,the distance between their centers $d$ must satisfy the condition $|r_1 - r_2| < d < r_1 + r_2$.
For the first circle $x^2+y^2-10x+16=0$,the center $C_1$ is $(5, 0)$ and the radius $r_1 = \sqrt{5^2 + 0^2 - 16} = \sqrt{25-16} = 3$.
For the second circle $x^2+y^2=a^2$,the center $C_2$ is $(0, 0)$ and the radius $r_2 = |a|$.
The distance between centers $d = \sqrt{(5-0)^2 + (0-0)^2} = 5$.
Applying the condition $|r_1 - r_2| < d < r_1 + r_2$:
$|3 - |a|| < 5 < 3 + |a|$.
From $5 < 3 + |a|$,we get $|a| > 2$,which implies $a > 2$ or $a < -2$.
From $|3 - |a|| < 5$,we get $-5 < 3 - |a| < 5$,which simplifies to $-8 < -|a| < 2$,or $-2 < |a| < 8$.
Combining these,we get $2 < |a| < 8$. Since $a$ is a radius,$a > 0$,so $2 < a < 8$.

(A) Let $S_{1} = x^{2}+y^{2}-6x-8=0$ and $S_{2} = x^{2}+y^{2}-6=0$.
The equation of the family of circles passing through the intersection of $S_{1}$ and $S_{2}$ is given by $S_{1} + \lambda S_{2} = 0$.
$(x^{2}+y^{2}-6x-8) + \lambda(x^{2}+y^{2}-6) = 0 \quad \dots(i)$
Since the circle passes through the point $(1, 1)$,we substitute $x=1$ and $y=1$ into equation $(i)$:
$(1^{2}+1^{2}-6(1)-8) + \lambda(1^{2}+1^{2}-6) = 0$
$(1+1-6-8) + \lambda(1+1-6) = 0$
$-12 + \lambda(-4) = 0$
$-4\lambda = 12 \Rightarrow \lambda = -3$.
Substituting $\lambda = -3$ back into equation $(i)$:
$(x^{2}+y^{2}-6x-8) - 3(x^{2}+y^{2}-6) = 0$
$x^{2}+y^{2}-6x-8 - 3x^{2}-3y^{2}+18 = 0$
$-2x^{2}-2y^{2}-6x+10 = 0$
Dividing by $-2$,we get $x^{2}+y^{2}+3x-5 = 0$.

(A) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-6x-8y=0$ and the line $L: x+y-1=0$ is given by $S + \lambda L = 0$.
$x^2+y^2-6x-8y + \lambda(x+y-1) = 0$
$x^2+y^2 + (\lambda-6)x + (\lambda-8)y - \lambda = 0$.
Since $AB$ is a diameter,the center of this circle must lie on the line $x+y-1=0$.
The center of the circle is $(-\frac{\lambda-6}{2}, -\frac{\lambda-8}{2})$.
Substituting the center into the line equation:
$-\frac{\lambda-6}{2} - \frac{\lambda-8}{2} - 1 = 0$
$-(\lambda-6) - (\lambda-8) - 2 = 0$
$-\lambda + 6 - \lambda + 8 - 2 = 0$
$-2\lambda + 12 = 0 \implies \lambda = 6$.
Substituting $\lambda = 6$ into the circle equation:
$x^2+y^2 + (6-6)x + (6-8)y - 6 = 0$
$x^2+y^2-2y-6=0$.

(C) The equation of the family of curves passing through the intersection of the two ellipses is given by $S_1 + \lambda S_2 = 0$,where $S_1: x^{2}+2y^{2}-6x-12y+20=0$ and $S_2: 2x^{2}+y^{2}-10x-6y+15=0$.
$(x^{2}+2y^{2}-6x-12y+20) + \lambda(2x^{2}+y^{2}-10x-6y+15) = 0$
$(1+2\lambda)x^{2} + (2+\lambda)y^{2} - (6+10\lambda)x - (12+6\lambda)y + (20+15\lambda) = 0$
For this to represent a circle,the coefficients of $x^{2}$ and $y^{2}$ must be equal:
$1+2\lambda = 2+\lambda \Rightarrow \lambda = 1$
Substituting $\lambda = 1$ into the equation:
$3x^{2} + 3y^{2} - 16x - 18y + 35 = 0$
$x^{2} + y^{2} - \frac{16}{3}x - 6y + \frac{35}{3} = 0$
The centre of the circle is given by $\left(-\frac{g}{2}, -\frac{f}{2}\right)$,where $2g = -\frac{16}{3}$ and $2f = -6$.
Centre $= \left(\frac{16/3}{2}, \frac{6}{2}\right) = \left(\frac{8}{3}, 3\right)$.

(D) Let the equations of the ellipses be $S_{1} = x^{2}+2y^{2}-6x-12y+23=0$ and $S_{2} = 4x^{2}+2y^{2}-20x-12y+35=0$.
Any curve passing through the intersection of these two ellipses is given by $S_{1} + \lambda S_{2} = 0$.
$(x^{2}+2y^{2}-6x-12y+23) + \lambda(4x^{2}+2y^{2}-20x-12y+35) = 0$.
$(1+4\lambda)x^{2} + (2+2\lambda)y^{2} - (6+20\lambda)x - (12+12\lambda)y + (23+35\lambda) = 0$.
For this to be a circle,the coefficient of $x^{2}$ must equal the coefficient of $y^{2}$:
$1+4\lambda = 2+2\lambda$ $\Rightarrow 2\lambda = 1$ $\Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the equation:
$(1+2)x^{2} + (2+1)y^{2} - (6+10)x - (12+6)y + (23+17.5) = 0$.
$3x^{2} + 3y^{2} - 16x - 18y + 40.5 = 0$.
Dividing by $3$: $x^{2} + y^{2} - \frac{16}{3}x - 6y + 13.5 = 0$.
The centre $(a, b)$ is $(\frac{8}{3}, 3)$.
The radius $r$ is given by $r^{2} = g^{2} + f^{2} - c = (\frac{8}{3})^{2} + 3^{2} - 13.5 = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$.
Thus,$ab + 18r^{2} = (\frac{8}{3} \times 3) + 18(\frac{47}{18}) = 8 + 47 = 55$.

(A) The first circle is $C_1: (x+1)^2+(y+4)^2=r^2$ with center $O_1(-1, -4)$ and radius $r_1 = |r|$.
The second circle is $C_2: x^2+y^2-4x-2y-4=0$. Rewriting in standard form: $(x-2)^2+(y-1)^2 = 4+1+4 = 9$,so center $O_2(2, 1)$ and radius $r_2 = 3$.
For two circles to intersect at two distinct points,the distance between their centers $d = O_1O_2$ must satisfy $|r_1 - r_2| < d < r_1 + r_2$.
Calculate $d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$.
Thus,$|r - 3| < \sqrt{34} < |r| + 3$.
From $|r - 3| < \sqrt{34}$,we have $-\sqrt{34} < r - 3 < \sqrt{34}$,which implies $3 - \sqrt{34} < r < 3 + \sqrt{34}$.
From $\sqrt{34} < |r| + 3$,we have $|r| > \sqrt{34} - 3$. Since $\sqrt{34} \approx 5.83$,$\sqrt{34} - 3 > 0$,so $r > \sqrt{34} - 3$ or $r < -(\sqrt{34} - 3)$.
Since $r$ is a radius,$r > 0$. The intersection of $r \in (3 - \sqrt{34}, 3 + \sqrt{34})$ and $r > \sqrt{34} - 3$ is $r \in (\sqrt{34} - 3, \sqrt{34} + 3)$.
Thus,$\alpha = \sqrt{34} - 3$ and $\beta = \sqrt{34} + 3$.
$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25$.