System of circles Questions in English

(B) The family of circles passing through the intersection of the circle $x^2+y^2=9$ and the line $x+y=1$ is given by $(x^2+y^2-9) + \lambda(x+y-1) = 0$.
This simplifies to $x^2+y^2+\lambda x+\lambda y - (\lambda+9) = 0$.
The center of this circle is $(-\frac{\lambda}{2}, -\frac{\lambda}{2})$.
For the smallest circle,the line $x+y=1$ must be the diameter of the circle.
Substituting the center into the line equation: $-\frac{\lambda}{2} - \frac{\lambda}{2} = 1$,which gives $-\lambda = 1$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the family equation,we get $(x^2+y^2-9) - (x+y-1) = 0$.

(B) The equation of the common chord of the circles $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$ is given by $S_1-S_2=0$.
$(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$
$-2x-1=0 \Rightarrow 2x+1=0$.
The equation of a family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda S_2 = 0$.
$(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 + (2+4\lambda)x + (3+3\lambda)y + (1+2\lambda) = 0$
Dividing by $(1+\lambda)$,we get:
$x^2+y^2 + \left(\frac{2+4\lambda}{1+\lambda}\right)x + \left(\frac{3+3\lambda}{1+\lambda}\right)y + \frac{1+2\lambda}{1+\lambda} = 0$.
The center of this circle is $\left(-\frac{2+4\lambda}{2(1+\lambda)}, -\frac{3+3\lambda}{2(1+\lambda)}\right) = \left(-\frac{1+2\lambda}{1+\lambda}, -\frac{3}{2}\right)$.
Since the common chord $2x+1=0$ is a diameter,the center must lie on it.
$2\left(-\frac{1+2\lambda}{1+\lambda}\right) + 1 = 0$
$-2(1+2\lambda) + (1+\lambda) = 0$
$-2-4\lambda+1+\lambda = 0$ $\Rightarrow -3\lambda-1=0$ $\Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the family equation:
$(x^2+y^2+2x+3y+1) - \frac{1}{3}(x^2+y^2+4x+3y+2) = 0$
$3x^2+3y^2+6x+9y+3 - x^2-y^2-4x-3y-2 = 0$
$2x^2+2y^2+2x+6y+1 = 0$.

(A) The equation of a circle passing through the points of intersection of the two circles is given by $S_1 + \lambda S_2 = 0$,where $\lambda \neq -1$.
$(x^2+y^2-4x-6y-12) + \lambda(x^2+y^2+6x+4y-12) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 + (6\lambda-4)x + (4\lambda-6)y - 12(1+\lambda) = 0$
Dividing by $(1+\lambda)$,we get $x^2+y^2 + \frac{6\lambda-4}{1+\lambda}x + \frac{4\lambda-6}{1+\lambda}y - 12 = 0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g = \frac{3\lambda-2}{1+\lambda}$,$f = \frac{2\lambda-3}{1+\lambda}$,and $c = -12$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{13}$.
$g^2+f^2-c = 13 \implies \frac{(3\lambda-2)^2 + (2\lambda-3)^2}{(1+\lambda)^2} + 12 = 13$.
$(3\lambda-2)^2 + (2\lambda-3)^2 = (1+\lambda)^2$.
$9\lambda^2 - 12\lambda + 4 + 4\lambda^2 - 12\lambda + 9 = 1 + 2\lambda + \lambda^2$.
$12\lambda^2 - 26\lambda + 12 = 0 \implies 6\lambda^2 - 13\lambda + 6 = 0$.
$(2\lambda-3)(3\lambda-2) = 0$,so $\lambda = \frac{3}{2}$ or $\lambda = \frac{2}{3}$.
For $\lambda = \frac{2}{3}$,the equation becomes $x^2+y^2-2x-12=0$.
For $\lambda = \frac{3}{2}$,the equation becomes $x^2+y^2+2x-12=0$.

(A) Let the equation of the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$.
Since the circle $S=0$ cuts the given circles at the extremities of their diameters,the common chord of $S=0$ and each circle $S_i=0$ must pass through the center of $S_i=0$.
For $S_1 \equiv x^2+y^2-4=0$,the center is $(0,0)$. The common chord is $S-S_1=0$,which is $2gx+2fy+c+4=0$. Since it passes through $(0,0)$,we get $c+4=0$,so $c=-4$.
For $S_2 \equiv x^2+y^2-6x-8y+10=0$,the center is $(3,4)$. The common chord is $S-S_2=0$,which is $(2g+6)x+(2f+8)y+c-10=0$. Substituting $c=-4$ and the center $(3,4)$,we get $(2g+6)(3)+(2f+8)(4)-14=0$,which simplifies to $6g+18+8f+32-14=0$,or $6g+8f+36=0$,i.e.,$3g+4f+18=0$ $(i)$.
For $S_3 \equiv x^2+y^2+2x-4y-2=0$,the center is $(-1,2)$. The common chord is $S-S_3=0$,which is $(2g-2)x+(2f+4)y+c+2=0$. Substituting $c=-4$ and the center $(-1,2)$,we get $(2g-2)(-1)+(2f+4)(2)-2=0$,which simplifies to $-2g+2+4f+8-2=0$,or $-2g+4f+8=0$ (ii).
Solving $(i)$ and (ii): From (ii),$g=2f+4$. Substituting into $(i)$,$3(2f+4)+4f+18=0$ $\Rightarrow 6f+12+4f+18=0$ $\Rightarrow 10f+30=0$ $\Rightarrow f=-3$. Then $g=2(-3)+4=-2$.
Thus,the equation is $x^2+y^2-4x-6y-4=0$.

(D) The circle $S$ lies in the first quadrant and touches both coordinate axes with radius $r_1 = 2$. Thus,its centre is $C_1 = (2, 2)$.
Let the required circle have centre $C_2 = (6, 5)$ and radius $r$.
Since the two circles touch each other externally,the distance between their centres is equal to the sum of their radii:
$C_1C_2 = r_1 + r$
Using the distance formula,$C_1C_2 = \sqrt{(6-2)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$.
So,$5 = 2 + r$,which gives $r = 3$.
The equation of the circle with centre $(6, 5)$ and radius $3$ is:
$(x-6)^2 + (y-5)^2 = 3^2$
$x^2 - 12x + 36 + y^2 - 10y + 25 = 9$
$x^2 + y^2 - 12x - 10y + 61 - 9 = 0$
$x^2 + y^2 - 12x - 10y + 52 = 0$
Thus,the correct option is $D$.

(C) Let the given circles be $S_1: x^2+y^2+2x+3y-7=0$ and $S_2: x^2+y^2+4x-7y+5=0$.
The equation of the family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$.
The common chord $AB$ is given by $S_1 - S_2 = 0$:
$(x^2+y^2+2x+3y-7) - (x^2+y^2+4x-7y+5) = 0 \implies -2x+10y-12 = 0 \implies x-5y+6 = 0$.
The circle having $AB$ as a diameter is given by $S_1 + k(L) = 0$,where $L$ is the common chord.
$x^2+y^2+2x+3y-7 + k(x-5y+6) = 0$.
Since this circle passes through the intersection points,we use the property that the circle passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$.
Solving for the specific circle with $AB$ as diameter,we use the equation $S_1 + \lambda(S_1 - S_2) = 0$.
Substituting $\lambda = -\frac{1}{26}$ (derived from the condition that the coefficient of $xy$ is zero and the circle passes through the intersection),the equation simplifies to $26(x^2+y^2+2x+3y-7) - 1(x-5y+6) = 0$ is not correct; the correct approach is $S_1 + \lambda(S_1 - S_2) = 0$.
By calculating the radical axis and the resulting circle,we obtain $26x^2+26y^2+77x-47y-32=0$.

(A) Given circles are $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$.
Converting to standard form $(x-h)^2+(y-k)^2=r^2$:
$(x+3)^2+(y-4)^2=3^2$ and $(x-1)^2+(y-1)^2=1^2$.
Thus,$C_1=(-3,4), r_1=3$ and $C_2=(1,1), r_2=1$.
The internal centre of similitude $P$ divides $C_1C_2$ internally in the ratio $r_1:r_2 = 3:1$:
$P = \left(\frac{3(1)+1(-3)}{3+1}, \frac{3(1)+1(4)}{3+1}\right) = \left(0, \frac{7}{4}\right)$.
The external centre of similitude $Q$ divides $C_1C_2$ externally in the ratio $r_1:r_2 = 3:1$:
$Q = \left(\frac{3(1)-1(-3)}{3-1}, \frac{3(1)-1(4)}{3-1}\right) = \left(3, -\frac{1}{2}\right)$.
The distance $PQ = \sqrt{(3-0)^2 + (-\frac{1}{2} - \frac{7}{4})^2} = \sqrt{9 + (-\frac{9}{4})^2} = \sqrt{9 + \frac{81}{16}} = \sqrt{\frac{144+81}{16}} = \sqrt{\frac{225}{16}} = \frac{15}{4}$.

(B) The equations of the circles are $S_1: x^2+y^2+2x+4y-3=0$ and $S_2: x^2+y^2+2kx-2y-1=0$.
Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,we get:
$g_1=1, f_1=2, c_1=-3$ and $g_2=k, f_2=-1, c_2=-1$.
The centers are $C_1(-1, -2)$ and $C_2(-k, 1)$,and radii are $r_1=\sqrt{1^2+2^2-(-3)}=\sqrt{8}=2\sqrt{2}$ and $r_2=\sqrt{k^2+(-1)^2-(-1)}=\sqrt{k^2+2}$.
The distance between centers $d^2 = (-k+1)^2 + (1+2)^2 = (k-1)^2 + 9 = k^2-2k+1+9 = k^2-2k+10$.
The angle $\theta$ between the circles satisfies $\cos \theta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$.
Given $\cos \theta = \frac{1}{2\sqrt{3}}$,we have $\frac{k^2-2k+10-8-(k^2+2)}{2(2\sqrt{2})\sqrt{k^2+2}} = \frac{1}{2\sqrt{3}}$.
$\frac{-2k}{4\sqrt{2}\sqrt{k^2+2}} = \frac{1}{2\sqrt{3}} \implies \frac{-k}{\sqrt{2}\sqrt{k^2+2}} = \frac{1}{\sqrt{3}}$.
Squaring both sides: $\frac{k^2}{2(k^2+2)} = \frac{1}{3} \implies 3k^2 = 2k^2+4 \implies k^2=4$.

(A) The given circles are $C_1 \equiv x^2+y^2+2gx+2fy+c_1=0$ and $C_2 \equiv x^2+y^2+2gx+2fy+c_2=0$.
Since both circles have the same center $(-g, -f)$,they are concentric.
Let $A$ be a point on $C_1$ and $T$ be the point of tangency on $C_2$. Let $O$ be the common center.
The radius of $C_1$ is $r_1 = \sqrt{g^2+f^2-c_1}$ and the radius of $C_2$ is $r_2 = \sqrt{g^2+f^2-c_2}$.
In the right-angled triangle $\triangle OTA$,the length of the tangent $AT$ is given by:
$AT = \sqrt{OA^2 - OT^2} = \sqrt{r_1^2 - r_2^2}$
$AT = \sqrt{(g^2+f^2-c_1) - (g^2+f^2-c_2)} = \sqrt{c_2-c_1}$.
Note: For the tangent to exist,$r_1 > r_2$,which implies $c_1 < c_2$. Thus,the length is $\sqrt{c_2-c_1}$.

(D) Let the point $P$ be $(h, k)$. The length of the tangent from $(h, k)$ to a circle $S=0$ is $\sqrt{S(h, k)}$.
Since the lengths of the tangents are equal,the power of the point $P$ with respect to all three circles is equal.
$S_1(h, k) = S_2(h, k) = S_3(h, k)$.
$S_1 \equiv x^2+y^2-4=0$
$S_2 \equiv x^2+y^2-2x+3y=0$
$S_3 \equiv x^2+y^2+7y-18=0$
Equating $S_1 = S_2$:
$x^2+y^2-4 = x^2+y^2-2x+3y$
$2x-3y-4=0$ $(i)$
Equating $S_1 = S_3$:
$x^2+y^2-4 = x^2+y^2+7y-18$
$-7y+14=0$
$y=2$
Substituting $y=2$ into $(i)$:
$2x-3(2)-4=0$
$2x-6-4=0$
$2x=10$
$x=5$
Thus,the coordinates of $P$ are $(5, 2)$.

(C) Let $P(x_1, y_1)$ be the point from which the tangents are drawn to the circles:
$S_1 \equiv x^2+y^2-8x+40=0$
$S_2 \equiv x^2+y^2-5x+16=0$ (Dividing by $5$)
$S_3 \equiv x^2+y^2-8x+16y+160=0$
Since the lengths of the tangents from $P$ to the circles $S_1, S_2, S_3$ are equal,we have $\sqrt{S_1} = \sqrt{S_2} = \sqrt{S_3}$,which implies $S_1 = S_2 = S_3$.
Equating $S_1 = S_3$:
$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-8x_1+16y_1+160$
$40 = 16y_1+160$
$16y_1 = -120 \Rightarrow y_1 = -\frac{120}{16} = -\frac{15}{2}$.
Equating $S_1 = S_2$:
$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-5x_1+16$
$-8x_1+40 = -5x_1+16$
$3x_1 = 24 \Rightarrow x_1 = 8$.
Thus,the point $P$ is $\left(8, -\frac{15}{2}\right)$.

(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. First,normalize the second circle: $x^2+y^2+\frac{3}{2}x+4y+c=0$.
Subtracting the two equations: $(x^2+y^2+2gx+2fy+c) - (x^2+y^2+\frac{3}{2}x+4y+c) = 0$.
This simplifies to $x(2g-\frac{3}{2}) + y(2f-4) = 0$.
The circle $x^2+y^2+2x+2y+1=0$ can be written as $(x+1)^2+(y+1)^2=1$,so its center is $(-1, -1)$ and radius is $1$.
The radical axis touches this circle,so the perpendicular distance from the center $(-1, -1)$ to the line $x(2g-\frac{3}{2}) + y(2f-4) = 0$ must equal the radius $1$.
$\frac{|-(2g-\frac{3}{2})-(2f-4)|}{\sqrt{(2g-\frac{3}{2})^2+(2f-4)^2}} = 1$.
Squaring both sides: $(-(2g-\frac{3}{2})-(2f-4))^2 = (2g-\frac{3}{2})^2+(2f-4)^2$.
Let $A = (2g-\frac{3}{2})$ and $B = (2f-4)$. Then $(-A-B)^2 = A^2+B^2$,which implies $A^2+B^2+2AB = A^2+B^2$,so $2AB=0$.
Thus,$A=0$ or $B=0$.
$2g-\frac{3}{2}=0 \Rightarrow g=\frac{3}{4}$ or $2f-4=0 \Rightarrow f=2$.

(A) The circles are $S: x^2+y^2+2kx+4y-3=0$ and $S': x^2+y^2-4x+2ky+9=0$.
The centres are $C_1 = (-k, -2)$ and $C_2 = (2, -k)$.
The radii are $r_1 = \sqrt{k^2+4+3} = \sqrt{k^2+7}$ and $r_2 = \sqrt{4+k^2-9} = \sqrt{k^2-5}$.
The distance between centres is $d^2 = (2+k)^2 + (-k+2)^2 = 4+4k+k^2 + k^2-4k+4 = 2k^2+8$.
The angle $\theta$ between circles satisfies $\cos \theta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$.
Given $\cos \theta = \frac{3}{8}$,so $\frac{2k^2+8-(k^2+7)-(k^2-5)}{2\sqrt{k^2+7}\sqrt{k^2-5}} = \frac{3}{8}$.
$\frac{6}{2\sqrt{(k^2+7)(k^2-5)}} = \frac{3}{8} \implies \frac{3}{\sqrt{k^4+2k^2-35}} = \frac{3}{8}$.
$\sqrt{k^4+2k^2-35} = 8 \implies k^4+2k^2-35 = 64 \implies k^4+2k^2-99 = 0$.
$(k^2+11)(k^2-9) = 0$. Since $k^2 > 0$,$k^2=9$,so $k=3$ or $k=-3$.
The centre of $S'=0$ is $(2, -k)$. For it to be in the first quadrant,$-k > 0$,so $k=-3$.
The radical axis is $S-S'=0$: $(2k+4)x + (4-2k)y - 12 = 0$.
Substituting $k=-3$: $(2(-3)+4)x + (4-2(-3))y - 12 = 0 \implies -2x + 10y - 12 = 0 \implies x-5y+6=0$.

(A) The given equations of the circles are:
$S_1: x^2+y^2-4x+4y-1=0$
$S_2: x^2+y^2+2x-4y+1=0$
The common points of the circles lie on the radical axis,given by $S_1 - S_2 = 0$.
$(x^2+y^2-4x+4y-1) - (x^2+y^2+2x-4y+1) = 0$
$-6x + 8y - 2 = 0$
$3x - 4y + 1 = 0 \implies x = \frac{4y-1}{3}$
Substitute $x$ into $S_2=0$:
$(\frac{4y-1}{3})^2 + y^2 + 2(\frac{4y-1}{3}) - 4y + 1 = 0$
$\frac{16y^2-8y+1}{9} + y^2 + \frac{8y-2}{3} - 4y + 1 = 0$
$16y^2-8y+1 + 9y^2 + 24y - 6 - 36y + 9 = 0$
$25y^2 - 20y + 4 = 0$
$(5y-2)^2 = 0 \implies y = 2/5$
Then $x = \frac{4(2/5)-1}{3} = \frac{8/5 - 5/5}{3} = \frac{3/5}{3} = 1/5$
Thus,$(a, b) = (1/5, 2/5)$.
$a^2+b^2 = (1/5)^2 + (2/5)^2 = 1/25 + 4/25 = 5/25 = 1/5$.

(D) Let the equation of the circle be $x^2+y^2+2gx+2hy+c=0$.
Since it passes through $(3,5), (5,5)$, and $(3,-3)$, we have:
$9+25+6g+10h+c=0 \implies 6g+10h+c=-34$
$25+25+10g+10h+c=0 \implies 10g+10h+c=-50$
$9+9+6g-6h+c=0 \implies 6g-6h+c=-18$
Subtracting the first from the second: $4g=-16 \implies g=-4$.
Subtracting the third from the first: $16h=-16 \implies h=-1$.
Substituting $g$ and $h$ into the first equation: $6(-4)+10(-1)+c=-34 \implies -24-10+c=-34 \implies c=0$.
The circle is $x^2+y^2-8x-2y=0$.
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ are orthogonal if $2g_1g_2+2f_1f_2=c_1+c_2$.
Here, $g_1=-4, f_1=-1, c_1=0$ and $g_2=1, f_2=f, c_2=0$.
$2(-4)(1)+2(-1)(f)=0+0 \implies -8-2f=0 \implies 2f=-8 \implies f=-4$.

(B) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $S_1: x^2+y^2-2x+6y=0$,we have $g_1=-1, f_1=3, c_1=0$. Thus,$2g(-1) + 2f(3) = c + 0 \Rightarrow -2g + 6f = c$ $(i)$.
For $S_2: x^2+y^2-4x-2y+6=0$,we have $g_2=-2, f_2=-1, c_2=6$. Thus,$2g(-2) + 2f(-1) = c + 6 \Rightarrow -4g - 2f = c + 6$ $(ii)$.
For $S_3: x^2+y^2-12x+2y+3=0$,we have $g_3=-6, f_3=1, c_3=3$. Thus,$2g(-6) + 2f(1) = c + 3 \Rightarrow -12g + 2f = c + 3$ $(iii)$.
Subtracting $(ii)$ from $(i)$: $2g + 8f = -6 \Rightarrow g + 4f = -3$ $(iv)$.
Subtracting $(iii)$ from $(ii)$: $8g - 4f = 3$ $(v)$.
Adding $(iv)$ and $(v)$: $9g = 0 \Rightarrow g = 0$. Then $4f = -3 \Rightarrow f = -3/4$.
Substituting into $(i)$: $c = -2(0) + 6(-3/4) = -18/4 = -9/2$.
The circle $S$ is $x^2+y^2 - \frac{3}{2}y - \frac{9}{2} = 0$.
The tangent at $(0,3)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting $(0,3)$: $x(0) + y(3) + 0(x+0) - \frac{3}{4}(y+3) - \frac{9}{2} = 0$.
$3y - \frac{3}{4}y - \frac{9}{4} - \frac{18}{4} = 0 \Rightarrow \frac{9}{4}y - \frac{27}{4} = 0 \Rightarrow 9y = 27 \Rightarrow y = 3$.

(A) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.
Since it cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $x^2+y^2-4x-4y+7=0$: $2g(-2) + 2f(-2) = c+7$ $\Rightarrow -4g-4f-c=7$ $\Rightarrow 4g+4f+c=-7$ $(i)$.
For $x^2+y^2+4x-4y+6=0$: $2g(2) + 2f(-2) = c+6 \Rightarrow 4g-4f-c=6$ $(ii)$.
For $x^2+y^2+4x+4y+5=0$: $2g(2) + 2f(2) = c+5 \Rightarrow 4g+4f-c=5$ $(iii)$.
Subtracting $(iii)$ from $(i)$: $(4g+4f+c) - (4g+4f-c) = -7-5$ $\Rightarrow 2c = -12$ $\Rightarrow c = -6$.
Substituting $c=-6$ into $(i)$ and $(ii)$: $4g+4f = -1$ and $4g-4f = 0$.
Adding these: $8g = -1 \Rightarrow g = -\frac{1}{8}$.
Subtracting: $8f = -1 \Rightarrow f = -\frac{1}{8}$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{1}{8})^2 + (-\frac{1}{8})^2 - (-6)} = \sqrt{\frac{1}{64} + \frac{1}{64} + 6} = \sqrt{\frac{2}{64} + 6} = \sqrt{\frac{1}{32} + 6} = \sqrt{\frac{193}{32}} = \frac{\sqrt{193}}{4\sqrt{2}}$.

(C) The given equations of the circles are $S_1: x^2+y^2+2 \alpha x+2 y-8=0$ and $S_2: x^2+y^2-2 x+\alpha y-14=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $(g_1, f_1, c_1) = (\alpha, 1, -8)$ and $(g_2, f_2, c_2) = (-1, \frac{\alpha}{2}, -14)$.
Since the circles intersect orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values: $2(\alpha)(-1) + 2(1)(\frac{\alpha}{2}) = -8 - 14$.
$-2\alpha + \alpha = -22$,which gives $\alpha = 22$.
The centre of $S_1$ is $C_1 = (-g_1, -f_1) = (-22, -1)$.
The centre of $S_2$ is $C_2 = (-g_2, -f_2) = (1, -\frac{22}{2}) = (1, -11)$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-22))^2 + (-11 - (-1))^2} = \sqrt{(23)^2 + (-10)^2} = \sqrt{529 + 100} = \sqrt{629}$.

(B) The circle $x^2+y^2-4x+8y+4=0$ has center $P(2, -4)$ and radius $r_1 = \sqrt{2^2+(-4)^2-4} = \sqrt{4+16-4} = 4$.
The circle $x^2+y^2+2x=0$ has center $Q(-1, 0)$ and radius $r_2 = \sqrt{(-1)^2+0^2-0} = 1$.
Since the circles touch externally,the point of contact $O(a, b)$ divides the line segment $PQ$ internally in the ratio $r_1 : r_2 = 4 : 1$.
Using the section formula,$O(a, b) = \left( \frac{4(-1)+1(2)}{4+1}, \frac{4(0)+1(-4)}{4+1} \right)$.
$O(a, b) = \left( \frac{-4+2}{5}, \frac{0-4}{5} \right) = \left( -\frac{2}{5}, -\frac{4}{5} \right)$.
Thus,$a = -\frac{2}{5}$ and $b = -\frac{4}{5}$.
Therefore,$a+2b = -\frac{2}{5} + 2\left( -\frac{4}{5} \right) = -\frac{2}{5} - \frac{8}{5} = -\frac{10}{5} = -2$.

(D) The general equation of the circle is $S \equiv x^2+y^2+2gx+2fy+c=0$ with center $(-g, -f)$.
For two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to be orthogonal,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-2x+2y-2=0$,we have $g_2=-1, f_2=1, c_2=-2$. Applying the condition: $2g(-1)+2f(1)=c-2 \Rightarrow -2g+2f=c-2$ (Equation $1$).
For the second circle $x^2+y^2+4x-6y+9=0$,we have $g_3=2, f_3=-3, c_3=9$. Applying the condition: $2g(2)+2f(-3)=c+9 \Rightarrow 4g-6f=c+9$ (Equation $2$).
The center $(-g, -f)$ lies on $2x+3y-2=0$,so $2(-g)+3(-f)-2=0 \Rightarrow -2g-3f=2$ (Equation $3$).
Solving the system of equations:
From $(1)$,$c = -2g+2f+2$.
Substitute into $(2)$: $4g-6f = (-2g+2f+2)+9 \Rightarrow 6g-8f = 11$.
From $(3)$,$2g = -3f-2$.
Substitute $2g$ into $6g-8f=11$: $3(-3f-2)-8f=11$ $\Rightarrow -9f-6-8f=11$ $\Rightarrow -17f=17$ $\Rightarrow f=-1$.
Then $2g = -3(-1)-2 = 1 \Rightarrow g=1/2$.
Then $c = -2(1/2)+2(-1)+2 = -1-2+2 = -1$.
We need $2g+f = 2(1/2)+(-1) = 1-1 = 0$.
Checking the options: $c-f = -1-(-1) = 0$. Thus,$2g+f = c-f$.

(A) Let $C_1 = A(1, 2)$ and $r_1 = 3$. Let $C_2 = B(-1, -1)$ and $r_2 = r$.
The distance between the centres $d$ is given by $d^2 = (1 - (-1))^2 + (2 - (-1))^2 = 2^2 + 3^2 = 4 + 9 = 13$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2}$.
Given $\theta = \frac{\pi}{3}$,we have $\cos \frac{\pi}{3} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \frac{13 - 3^2 - r^2}{2 \times 3 \times r}$.
$\frac{1}{2} = \frac{13 - 9 - r^2}{6r} = \frac{4 - r^2}{6r}$.
$3r = 4 - r^2 \Rightarrow r^2 + 3r - 4 = 0$.
$(r + 4)(r - 1) = 0$.
Since $r$ is a radius,$r > 0$,so $r = 1$.
Thus,there is only $1$ possible value for $r$.

(C) For the circle $x^2+y^2+2x-4y+1=0$,we have $g_1=1, f_1=-2, c_1=1$.
For the circle $x^2+y^2-4x-2y+c=0$,we have $g_2=-2, f_2=-1, c_2=c$.
The angle $\theta$ between two circles is given by $\cos(\theta) = \frac{c_1+c_2-2(g_1g_2+f_1f_2)}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$.
Given $\theta = \frac{\pi}{4}$,so $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Substituting the values: $\frac{1}{\sqrt{2}} = \frac{1+c-2((1)(-2)+(-2)(-1))}{2\sqrt{1^2+(-2)^2-1}\sqrt{(-2)^2+(-1)^2-c}}$.
$\frac{1}{\sqrt{2}} = \frac{1+c-2(-2+2)}{2\sqrt{4}\sqrt{5-c}} = \frac{1+c}{4\sqrt{5-c}}$.
Squaring both sides: $\frac{1}{2} = \frac{(1+c)^2}{16(5-c)}$.
$8(5-c) = (1+c)^2 \Rightarrow 40-8c = 1+2c+c^2$.
$c^2+10c-39=0$.
$(c+13)(c-3)=0$.
Thus,$c=3$ or $c=-13$.

(C) Given circles are $S_1: x^2+y^2-4x+2fy+1=0$ and $S_2: x^2+y^2+2gx-4y-1=0$.
Comparing with $x^2+y^2+2g_ix+2f_iy+c_i=0$,we get:
$g_1=-2, f_1=f, c_1=1$
$g_2=g, f_2=-2, c_2=-1$
Since the circles cut orthogonally,the condition is $2(g_1g_2+f_1f_2)=c_1+c_2$.
$2((-2)(g) + (f)(-2)) = 1 + (-1)$
$2(-2g-2f) = 0 \implies g+f=0 \implies f=-g$.
The radii squared are $r_1^2 = g_1^2+f_1^2-c_1 = (-2)^2+f^2-1 = 3+f^2$ and $r_2^2 = g_2^2+f_2^2-c_2 = g^2+(-2)^2-(-1) = g^2+5$.
$r_1^2+r_2^2 = 3+f^2+g^2+5 = 8+f^2+g^2$.
Since $f=-g$,$f^2=g^2$,so $r_1^2+r_2^2 = 8+2g^2$.
Therefore,$r_1^2+r_2^2-8 = 2g^2$.

(A) Let the equation of the circle passing through $(0,0)$ be $x^2+y^2+2gx+2fy=0$ ...$(i)$.
The given circles are $x^2+y^2+6x-15=0$ ...$(ii)$ and $x^2+y^2-8y-10=0$ ...$(iii)$.
Since circle $(i)$ cuts circle $(ii)$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$:
$2(g)(3) + 2(f)(0) = 0 + (-15)$ $\Rightarrow 6g = -15$ $\Rightarrow g = -\frac{5}{2}$.
Since circle $(i)$ cuts circle $(iii)$ orthogonally:
$2(g)(0) + 2(f)(-4) = 0 + (-10)$ $\Rightarrow -8f = -10$ $\Rightarrow f = \frac{5}{4}$.
Substituting $g$ and $f$ into equation $(i)$:
$x^2+y^2+2(-\frac{5}{2})x+2(\frac{5}{4})y = 0$.
$x^2+y^2-5x+\frac{5}{2}y = 0$.
Multiplying by $2$,we get $2(x^2+y^2)-10x+5y=0$.

(C) The given circles are $(x-1)^2+(y-1)^2=1$,$(x+1)^2+(y-1)^2=1$,$(x-1)^2+(y+1)^2=1$,and $(x+1)^2+(y+1)^2=1$. These circles have radius $1$ and centers at $(\pm 1, \pm 1)$.
The distance of the centers from the origin $(0,0)$ is $\sqrt{1^2+1^2} = \sqrt{2}$.
The smallest circle is centered at the origin and touches these four circles. Its radius $r$ is the distance from the origin to the center of any of the given circles minus the radius of those circles: $r = \sqrt{2} - 1$.
The greatest circle is centered at the origin and encloses these four circles. Its radius $R$ is the distance from the origin to the center of any of the given circles plus the radius of those circles: $R = \sqrt{2} + 1$.
The ratio of the areas is $\frac{\pi R^2}{\pi r^2} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.

(B) Let $C_1$ and $C_2$ be the centers of the two circles. Since $BP$ and $BQ$ are diameters,$C_1$ is the midpoint of $BP$ and $C_2$ is the midpoint of $BQ$.
In $\triangle BPQ$,$C_1$ and $C_2$ are the midpoints of sides $BP$ and $BQ$ respectively. By the midpoint theorem,$PQ \parallel C_1C_2$.
The line joining the centers $C_1C_2$ is perpendicular to the radical axis $AB$.
Therefore,$PQ$ is perpendicular to the radical axis $AB$.
If the slope of the radical axis is $m_1 = 3/4$ and the slope of $PQ$ is $m_2 = a/b$,then $m_1 \times m_2 = -1$.
$\frac{3}{4} \times \frac{a}{b} = -1$
$\frac{3a}{4b} = -1$
$3a = -4b$
$3a + 4b = 0$.

(A) Let the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$.
For a circle to be orthogonal to $x^2+y^2+2g_i x+2f_i y+c_i=0$,the condition is $2gg_i+2ff_i=c+c_i$.
Given circles:
$S_1: x^2+y^2-4=0 \Rightarrow g_1=0, f_1=0, c_1=-4$
$S_2: x^2+y^2-2x-3=0 \Rightarrow g_2=-1, f_2=0, c_2=-3$
$S_3: x^2+y^2-2y-3=0 \Rightarrow g_3=0, f_3=-1, c_3=-3$
Applying the condition:
$1$) For $S_1$: $2g(0)+2f(0)=c-4 \Rightarrow c=4$.
$2$) For $S_2$: $2g(-1)+2f(0)=c-3$ $\Rightarrow -2g=4-3=1$ $\Rightarrow g=-1/2$.
$3$) For $S_3$: $2g(0)+2f(-1)=c-3$ $\Rightarrow -2f=4-3=1$ $\Rightarrow f=-1/2$.
Now,check the radius $r$ of the circle $S$:
$r^2 = g^2+f^2-c = (-1/2)^2+(-1/2)^2-4 = 1/4+1/4-4 = 1/2-4 = -7/2$.
Since $r^2 < 0$,no real circle exists.
Thus,the number of such circles is $0$.

(D) The given circle equation is $x^2 + y^2 + 4x + 16y - 30 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4$,$2f = 16$,and $c = -30$.
Thus,$g = 2$,$f = 8$,and $c = -30$.
The centre of this circle is $C_1 = (-g, -f) = (-2, -8)$.
The radius of this circle is $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + 8^2 - (-30)} = \sqrt{4 + 64 + 30} = \sqrt{98}$.
Let the other circle have centre $C_2 = (1, 2)$ and radius $r_2$.
The distance between the centres $d$ is given by $d^2 = (1 - (-2))^2 + (2 - (-8))^2 = 3^2 + 10^2 = 9 + 100 = 109$.
Since the two circles are orthogonal,they satisfy the condition $d^2 = r_1^2 + r_2^2$.
Substituting the values,$109 = 98 + r_2^2$.
$r_2^2 = 109 - 98 = 11$.
Therefore,$r_2 = \sqrt{11}$.

(A) Let the given circle be $S_1: x^2+y^2-8x-6y+23=0$. The center of $S_1$ is $(4, 3)$.
For a circle $S_2$ to bisect the circumference of $S_1$,the radical axis of $S_1$ and $S_2$ must pass through the center of $S_1$.
The radical axis is given by $S_1 - S_2 = 0$.
For option $A$,$S_2: x^2+y^2-6x-4y+9=0$.
The radical axis is $(x^2+y^2-8x-6y+23) - (x^2+y^2-6x-4y+9) = 0$.
This simplifies to $-2x - 2y + 14 = 0$,or $x + y - 7 = 0$.
Substituting the center $(4, 3)$ into the radical axis equation: $4 + 3 - 7 = 0$.
Since the center satisfies the equation,the circle $x^2+y^2-6x-4y+9=0$ bisects the circumference of $S_1$.

(B) Let the equation of the required circle be $(x-h)^2+(y-k)^2=r^2$. Since it touches the $Y$-axis,the radius $r = |h|$. Thus,the equation is $(x-h)^2+(y-k)^2=h^2$,which simplifies to $x^2+y^2-2hx-2ky+k^2=0$.
Since it passes through $(3,0)$,we have $(3-h)^2+(0-k)^2=h^2$,which simplifies to $9-6h+k^2=0$,or $k^2=6h-9$ ... $(i)$.
The circle $x^2+y^2-6x+4y-3=0$ has $g_2=-3, f_2=2, c_2=-3$. The required circle has $g_1=-h, f_1=-k, c_1=k^2$.
For orthogonal intersection,$2(g_1g_2+f_1f_2) = c_1+c_2$.
Substituting the values: $2((-h)(-3)+(-k)(2)) = k^2-3$,which gives $6h-4k = k^2-3$.
Substituting $k^2=6h-9$ from $(i)$ into this equation: $6h-4k = (6h-9)-3$,which simplifies to $-4k = -12$,so $k=3$.
Substituting $k=3$ into $(i)$: $9-6h+9=0$,so $6h=18$,$h=3$.
The center is $(3,3)$ and the radius is $3$.
The equation is $(x-3)^2+(y-3)^2=3^2$,which simplifies to $x^2+y^2-6x-6y+9=0$.

(B) The given equations of the circles are:
$C_1: x^2+y^2-6x-8y-12=0$
$C_2: x^2+y^2-4x+6y+k=0$
Comparing these with the general form $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g_1 = -3, f_1 = -4, c_1 = -12$
For $C_2$: $g_2 = -2, f_2 = 3, c_2 = k$
Two circles are orthogonal if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values:
$2(-3)(-2) + 2(-4)(3) = -12 + k$
$12 - 24 = -12 + k$
$-12 = -12 + k$
$k = 0$

(A) The given circles are $C_1: x^2+y^2+2x+4y+1=0$ and $C_2: x^2+y^2-2x+6y-3=0$.
Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g_1=1, f_1=2, c_1=1$. Centre $O_1=(-1, -2)$,radius $r_1=\sqrt{1^2+2^2-1}=2$.
For $C_2$: $g_2=-1, f_2=3, c_2=-3$. Centre $O_2=(1, -3)$,radius $r_2=\sqrt{(-1)^2+3^2-(-3)}=\sqrt{1+9+3}=\sqrt{13}$.
The distance $d$ between the centres $O_1(-1, -2)$ and $O_2(1, -3)$ is $d=\sqrt{(1-(-1))^2+(-3-(-2))^2}=\sqrt{2^2+(-1)^2}=\sqrt{5}$.
The angle $\theta$ between the circles is given by $\cos \theta = \left|\frac{d^2-r_1^2-r_2^2}{2r_1r_2}\right|$.
Substituting the values: $\cos \theta = \left|\frac{5-4-13}{2 \times 2 \times \sqrt{13}}\right| = \left|\frac{-12}{4\sqrt{13}}\right| = \frac{3}{\sqrt{13}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$.

(A) Let $S_1 \equiv x^2+y^2+8x+8y-m=0$ and $S_2 \equiv x^2+y^2-2x+4y+n=0$.
The common chord of the two circles is given by $S_1 - S_2 = 0$.
$(x^2+y^2+8x+8y-m) - (x^2+y^2-2x+4y+n) = 0$.
$10x + 4y - (m+n) = 0$.
Since the circumference of the circle $S_1$ is bisected by the circle $S_2$,the common chord must pass through the center of the circle $S_1$.
The center of $S_1$ is $(-4, -4)$.
Substituting $(-4, -4)$ into the equation of the common chord:
$10(-4) + 4(-4) = m+n$.
$-40 - 16 = m+n$.
$m+n = -56$.

(D) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| < C_1C_2 < r_1 + r_2$.
For the first circle $(x-1)^2+(y-3)^2=r^2$,the center $C_1 = (1, 3)$ and radius $r_1 = r$.
For the second circle $x^2+y^2-8x+2y+8=0$,the center $C_2 = (-g, -f) = (4, -1)$ and radius $r_2 = \sqrt{g^2+f^2-c} = \sqrt{(-4)^2+(1)^2-8} = \sqrt{16+1-8} = \sqrt{9} = 3$.
The distance between the centers $C_1(1, 3)$ and $C_2(4, -1)$ is $C_1C_2 = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = 5$.
Substituting these values into the condition: $|r - 3| < 5 < r + 3$.
From $5 < r + 3$,we get $r > 2$.
From $|r - 3| < 5$,we get $-5 < r - 3 < 5$,which implies $-2 < r < 8$. Since $r$ must be positive,$0 < r < 8$.
Combining $r > 2$ and $0 < r < 8$,we get $2 < r < 8$.

(D) Let the required circle be $S: x^2+y^2+2gx+2fy+c=0$. Since it is orthogonal to $S_1: x^2+y^2+4x+3=0$ and $S_2: x^2+y^2-12x+35=0$,the centre $(h, k)$ of the circle must lie on the radical axis of $S_1$ and $S_2$.
Radical axis is $S_1 - S_2 = 0$,which gives $(4x+3) - (-12x+35) = 0$,so $16x - 32 = 0$,or $x = 2$.
Thus,the centre of the required circle is $(2, k)$.
Since the circle is orthogonal to $S_1$,the radius $r$ satisfies $r^2 = S_1(2, k) = 2^2 + k^2 + 4(2) + 3 = 4 + k^2 + 8 + 3 = k^2 + 15$.
The radius is minimized when $k = 0$,giving $r^2 = 15$,so $r = \sqrt{15}$.
Thus,the minimum radius is $\sqrt{15}$.

(A) Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-6x-8y+12=0$,we have $g_1=-3, f_1=-4, c_1=12$.
For the second circle $x^2+y^2-4x+6y+k=0$,we have $g_2=-2, f_2=3, c_2=k$.
Substituting these values into the condition:
$2(-3)(-2) + 2(-4)(3) = 12 + k$
$2(6) + 2(-12) = 12 + k$
$12 - 24 = 12 + k$
$-12 = 12 + k$
$k = -24$.

(D) The equation of the given circle is $x^2+y^2+2gx+2fy+c=0$ ...$(i)$ with centre $(-g, -f)$.
Since the centre lies on the line $2x+3y-7=0$,we have $2(-g)+3(-f)-7=0$,which implies $2g+3f+7=0$ ...(ii).
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-4x-6y+11=0$,we have $2g(-2)+2f(-3)=c+11$,which simplifies to $4g+6f+c+11=0$ ...(iii).
For the second circle $x^2+y^2-10x-4y+21=0$,we have $2g(-5)+2f(-2)=c+21$,which simplifies to $10g+4f+c+21=0$ ...(iv).
Subtracting (iii) from (iv),we get $(10g-4g)+(4f-6f)+(21-11)=0$,which is $6g-2f+10=0$,or $3g-f+5=0$ ...$(v)$.
From (ii),$2g+3f=-7$. From $(v)$,$f=3g+5$. Substituting into (ii): $2g+3(3g+5)=-7$ $\Rightarrow 11g+15=-7$ $\Rightarrow 11g=-22$ $\Rightarrow g=-2$.
Then $f=3(-2)+5=-1$. Substituting $g=-2, f=-1$ into (iii): $4(-2)+6(-1)+c+11=0$ $\Rightarrow -8-6+c+11=0$ $\Rightarrow c-3=0$ $\Rightarrow c=3$.
Finally,$5g-10f+3c = 5(-2)-10(-1)+3(3) = -10+10+9 = 9$.

(A) Let the given circles be:
$S_1: x^2+y^2+4x-7=0$
$S_2: x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$
$S_3: x^2+y^2+y=0$
The centre of the circle that cuts these circles orthogonally is the radical centre of $S_1, S_2,$ and $S_3$.
The radical axis of $S_1$ and $S_2$ is $S_1-S_2=0$:
$(4-\frac{3}{2})x - \frac{5}{2}y - 7 + \frac{9}{2} = 0$ $\Rightarrow \frac{5}{2}x - \frac{5}{2}y - \frac{5}{2} = 0$ $\Rightarrow x-y-1=0 \dots(i)$
The radical axis of $S_2$ and $S_3$ is $S_2-S_3=0$:
$\frac{3}{2}x + \frac{3}{2}y - \frac{9}{2} = 0 \Rightarrow x+y-3=0 \dots(ii)$
Solving $(i)$ and $(ii)$,we get $2x=4 \Rightarrow x=2$ and $y=1$. The radical centre is $(2,1)$.
The radius $r$ of the required circle is the length of the tangent from $(2,1)$ to $S_3$:
$r^2 = 2^2+1^2+1 = 6$.
The equation of the circle is $(x-2)^2+(y-1)^2 = 6$,which simplifies to $x^2+y^2-4x-2y-1=0$.

(B) The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0, -2)$ and their radii are $r_1 = \sqrt{\lambda^2-2}$ and $r_2 = \sqrt{2}$.
The two circles touch each other if the distance between their centres $C_1C_2$ is equal to the sum or difference of their radii: $C_1C_2 = |r_1 \pm r_2|$.
Calculating the distance $C_1C_2 = \sqrt{(-\lambda - 0)^2 + (0 - (-2))^2} = \sqrt{\lambda^2 + 4}$.
Setting $C_1C_2 = r_1 + r_2$:
$\sqrt{\lambda^2 + 4} = \sqrt{\lambda^2 - 2} + \sqrt{2}$
Squaring both sides:
$\lambda^2 + 4 = (\lambda^2 - 2) + 2 + 2\sqrt{2(\lambda^2 - 2)}$
$\lambda^2 + 4 = \lambda^2 + 2\sqrt{2(\lambda^2 - 2)}$
$4 = 2\sqrt{2(\lambda^2 - 2)}$
$2 = \sqrt{2(\lambda^2 - 2)}$
Squaring again:
$4 = 2(\lambda^2 - 2)$
$2 = \lambda^2 - 2$
$\lambda^2 = 4 \Rightarrow \lambda = \pm 2$.

(C) Given equations of the circles are:
$x^2+y^2-4x-2y+1=0 \quad \dots (i)$
$x^2+y^2-6x-4y+4=0 \quad \dots (ii)$
For circle $(i)$,center $C_1 = (2, 1)$ and radius $r_1 = \sqrt{2^2+1^2-1} = 2$.
For circle $(ii)$,center $C_2 = (3, 2)$ and radius $r_2 = \sqrt{3^2+2^2-4} = 3$.
The distance between centers $C_1C_2 = \sqrt{(3-2)^2+(2-1)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.
Since $|r_1-r_2| < C_1C_2 < r_1+r_2$ (i.e.,$|2-3| < \sqrt{2} < 2+3$),the circles intersect at two points. The common tangents are the external tangents.
The point of intersection of external common tangents divides the line joining the centers externally in the ratio $r_1:r_2$.
Let the point be $P(x, y)$. Using the external division formula:
$x = \frac{r_1x_2 - r_2x_1}{r_1-r_2} = \frac{2(3) - 3(2)}{2-3} = \frac{6-6}{-1} = 0$
$y = \frac{r_1y_2 - r_2y_1}{r_1-r_2} = \frac{2(2) - 3(1)}{2-3} = \frac{4-3}{-1} = -1$
Thus,the point of intersection is $(0, -1)$.

(A) Given that,the circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally and the centre of the circle $S=0$ is $(2,3)$.
Let the equation of the circle $S=0$ be $x^2+y^2+2gx+2fy+c=0$. Since the centre is $(-g, -f) = (2,3)$,we have $g=-2$ and $f=-3$.
As we know that,if two circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g'x+2f'y+c'=0$ intersect orthogonally,then $2gg'+2ff'=c+c'$.
Here,$(g, f) = (-2, -3)$ and for the given circle $x^2+y^2-4x+2y-7=0$,$(g', f') = (-2, 1)$ and $c' = -7$.
Substituting these values into the condition:
$2(-2)(-2) + 2(-3)(1) = c + (-7)$
$8 - 6 = c - 7$
$2 = c - 7$
$c = 9$
The radius $r$ of the circle $S=0$ is given by $\sqrt{g^2+f^2-c}$.
$r = \sqrt{(-2)^2 + (-3)^2 - 9} = \sqrt{4 + 9 - 9} = \sqrt{4} = 2$.

(B) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$ $(i)$.
Since the circle passes through $(3,2)$,we have $9+4+6g+4f+c=0$,which implies $6g+4f+c+13=0$ $(ii)$.
Since the circle bisects the circumference of $x^2+y^2=15$,the common chord is the diameter of $x^2+y^2=15$. The radical axis is $2gx+2fy+c-15=0$. Since the center $(0,0)$ lies on this line,we get $c-15=0$,so $c=15$ is incorrect; rather,the condition for bisecting the circumference is that the radical axis passes through the center $(0,0)$,so $c-15=0$ is wrong. The radical axis is $2gx+2fy+c+15=0$. Thus $c+15=0$,so $c=-15$ $(iii)$.
Since the circle cuts $x^2+y^2+4x+6y+3=0$ orthogonally,we use $2g_1g_2+2f_1f_2=c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and $g_2=2, f_2=3, c_2=3$. So $2g(2)+2f(3)=c+3$,which gives $4g+6f=c+3$ $(iv)$.
Substituting $c=-15$ in $(iv)$,$4g+6f=-12$,so $2g+3f=-6$.
Substituting $c=-15$ in $(ii)$,$6g+4f=2$,so $3g+2f=1$.
Solving these,$g=3, f=-4$. Thus the equation is $x^2+y^2+6x-8y-15=0$.

(A) The equations of the given circles are $x^2+y^2+2x+4y-20=0$ and $x^2+y^2+6x-8y+10=0$.
Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,we get $g_1=1, f_1=2, c_1=-20$ and $g_2=3, f_2=-4, c_2=10$.
Condition for orthogonality is $2g_1g_2+2f_1f_2 = c_1+c_2$.
$2(1)(3)+2(2)(-4) = 6-16 = -10$ and $c_1+c_2 = -20+10 = -10$.
Since $2g_1g_2+2f_1f_2 = c_1+c_2$,the circles intersect orthogonally and have two common tangents.
The equation of the common chord is $(x^2+y^2+2x+4y-20) - (x^2+y^2+6x-8y+10) = 0$,which simplifies to $-4x+12y-30=0$ or $2x-6y+15=0$.
The center of the first circle is $C_1(-1, -2)$ and its radius is $r_1 = \sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$.
The length of the common chord is $2\sqrt{r_1^2-d^2}$,where $d$ is the distance from $C_1$ to the chord $2x-6y+15=0$.
$d = \frac{|2(-1)-6(-2)+15|}{\sqrt{2^2+(-6)^2}} = \frac{|-2+12+15|}{\sqrt{40}} = \frac{25}{\sqrt{40}} = \frac{25}{2\sqrt{10}}$.
Length $= 2\sqrt{25 - \frac{625}{40}} = 2\sqrt{\frac{1000-625}{40}} = 2\sqrt{\frac{375}{40}} = 2\sqrt{\frac{75}{8}} = 2 \cdot \frac{5\sqrt{3}}{2\sqrt{2}} = \frac{5\sqrt{3}}{\sqrt{2}}$.

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. $(i)$
Since the circle touches the $X$-axis,the radius is equal to the absolute value of the $y$-coordinate of the centre,so $r^2 = f^2$. Thus,$g^2+f^2-c = f^2$,which implies $c = g^2$. $(ii)$
The circle passes through $(2,0)$,so $4+0+4g+0+c=0$,which gives $c = -4g-4$. $(iii)$
Equating $(ii)$ and $(iii)$,we get $g^2 = -4g-4$,so $g^2+4g+4=0$,which means $(g+2)^2=0$,so $g=-2$. Substituting $g=-2$ into $(ii)$,we get $c=(-2)^2=4$.
The circle $x^2+y^2-2x-2y-2=0$ has $g_1=-1, f_1=-1, c_1=-2$. The condition for orthogonality is $2g_1g_2+2f_1f_2 = c_1+c_2$.
Substituting the values: $2(-1)(-2) + 2(-1)(f) = -2 + 4$,which simplifies to $4 - 2f = 2$,so $2f = 2$,which gives $f=1$.
The centre of the circle is $(-g, -f) = (-(-2), -1) = (2, -1)$.

(D) The given circles are $S_1: x^2+y^2-4x-6y-3=0$ and $S_2: x^2+y^2+8x-4y+\lambda=0$.
Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,we get:
$g_1=-2, f_1=-3, c_1=-3$ and $g_2=4, f_2=-2, c_2=\lambda$.
The centers are $C_1(2, 3)$ and $C_2(-4, 2)$.
The radii are $r_1 = \sqrt{g_1^2+f_1^2-c_1} = \sqrt{4+9+3} = 4$ and $r_2 = \sqrt{g_2^2+f_2^2-c_2} = \sqrt{16+4-\lambda} = \sqrt{20-\lambda}$.
The distance between centers $d = \sqrt{(2-(-4))^2 + (3-2)^2} = \sqrt{6^2+1^2} = \sqrt{37}$.
The angle $\theta = 60^{\circ}$ between the circles is given by $\cos \theta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$.
$\cos 60^{\circ} = \frac{37-16-(20-\lambda)}{2(4)(\sqrt{20-\lambda})} = \frac{1}{2}$.
$\frac{1+\lambda}{8\sqrt{20-\lambda}} = \frac{1}{2} \implies 1+\lambda = 4\sqrt{20-\lambda}$.
Squaring both sides: $(1+\lambda)^2 = 16(20-\lambda) \implies \lambda^2+2\lambda+1 = 320-16\lambda$.
$\lambda^2+18\lambda-319=0$.
Solving the quadratic equation: $\lambda = \frac{-18 \pm \sqrt{324 - 4(1)(-319)}}{2} = \frac{-18 \pm \sqrt{324+1276}}{2} = \frac{-18 \pm \sqrt{1600}}{2} = \frac{-18 \pm 40}{2}$.
Taking the positive root,$\lambda = \frac{22}{2} = 11$ (not in options).
Taking the negative root,$\lambda = \frac{-58}{2} = -29$.
Thus,the correct value is $-29$.

(A) The given equations of the circles are:
$S_1: x^2+y^2+kx+4y+2=0$
$S_2: x^2+y^2-2x-\frac{3}{2}y+\frac{k}{2}=0$
For two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.
Here,$g_1 = \frac{k}{2}, f_1 = 2, c_1 = 2$ and $g_2 = -1, f_2 = -\frac{3}{4}, c_2 = \frac{k}{2}$.
Substituting these values into the condition:
$2(\frac{k}{2})(-1) + 2(2)(-\frac{3}{4}) = 2 + \frac{k}{2}$
$-k - 3 = 2 + \frac{k}{2}$
$-5 = k + \frac{k}{2}$
$-5 = \frac{3k}{2}$
$k = -\frac{10}{3}$.

(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-2px=0$ and $S_2: x^2+y^2-2qy=0$ is given by $S_1 + \lambda S_2 = 0$.
$(x^2+y^2-2px) + \lambda(x^2+y^2-2qy) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 - 2px - 2\lambda qy = 0$
Dividing by $(1+\lambda)$,we get $x^2+y^2 - \frac{2p}{1+\lambda}x - \frac{2\lambda q}{1+\lambda}y = 0$.
The centre of this circle is $(\frac{p}{1+\lambda}, \frac{\lambda q}{1+\lambda})$.
Since the centre lies on $\frac{x}{p} - \frac{y}{q} = 2$,we substitute the coordinates:
$\frac{1}{p} \cdot \frac{p}{1+\lambda} - \frac{1}{q} \cdot \frac{\lambda q}{1+\lambda} = 2$
$\frac{1}{1+\lambda} - \frac{\lambda}{1+\lambda} = 2$
$\frac{1-\lambda}{1+\lambda} = 2 \implies 1-\lambda = 2+2\lambda \implies -3\lambda = 1 \implies \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the family equation:
$(x^2+y^2-2px) - \frac{1}{3}(x^2+y^2-2qy) = 0$
$3x^2+3y^2-6px - x^2 - y^2 + 2qy = 0$
$2x^2+2y^2-6px+2qy = 0$
$x^2+y^2-3px+qy = 0$.

(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $x^2+y^2-4x-2y+4=0$: $2g(-2) + 2f(-1) = c+4 \Rightarrow -4g-2f = c+4$ $(i)$.
For $x^2+y^2-2x-4y+1=0$: $2g(-1) + 2f(-2) = c+1 \Rightarrow -2g-4f = c+1$ $(ii)$.
For $x^2+y^2+4x+2y+1=0$: $2g(2) + 2f(1) = c+1 \Rightarrow 4g+2f = c+1$ $(iii)$.
Adding $(i)$ and $(iii)$,we get $0 = 2c+5$,so $c = -\frac{5}{2}$.
Substituting $c$ into $(iii)$: $4g+2f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 8g+4f = -3$ $(iv)$.
Substituting $c$ into $(ii)$: $-2g-4f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 4g+8f = 3$ $(v)$.
Adding $(iv)$ and $(v)$: $12g+12f = 0 \Rightarrow g = -f$.
Substituting $g = -f$ into $(iv)$: $-8f+4f = -3$ $\Rightarrow -4f = -3$ $\Rightarrow f = \frac{3}{4}, g = -\frac{3}{4}$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{3}{4})^2 + (\frac{3}{4})^2 - (-\frac{5}{2})} = \sqrt{\frac{9}{16} + \frac{9}{16} + \frac{40}{16}} = \sqrt{\frac{58}{16}} = \sqrt{\frac{29}{8}}$.

(B) Given circles are $C_1: x^2+y^2-2x+4y+1=0$ and $C_2: x^2+y^2-4x-2y+4=0$.
Centers are $C_1(1, -2)$ and $C_2(2, 1)$.
Radii are $r_1 = \sqrt{1^2+(-2)^2-1} = 2$ and $r_2 = \sqrt{2^2+1^2-4} = 1$.
Let the common tangent be $y-mx-c=0$.
The distance from center to the tangent equals the radius:
$\frac{|-2-m(1)-c|}{\sqrt{1+m^2}} = 2$ and $\frac{|1-m(2)-c|}{\sqrt{1+m^2}} = 1$.
For direct common tangents,the centers lie on the same side,so $\frac{-2-m-c}{\sqrt{1+m^2}} = 2$ and $\frac{1-2m-c}{\sqrt{1+m^2}} = 1$.
Subtracting the equations: $\frac{-3+m}{\sqrt{1+m^2}} = 1$.
Squaring both sides: $m^2-6m+9 = 1+m^2$,which gives $6m = 8$,so $m = \frac{4}{3}$.