Chord of contact of tangent and common chord Questions in English

(A) The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+2y+1) - (x^2+y^2+\alpha x+3y+2) = 0$
$(2-\alpha)x - y - 1 = 0$.
For the first circle $x^2+y^2+2x+2y+1=0$,the center is $(-1, -1)$ and the radius $r = \sqrt{(-1)^2 + (-1)^2 - 1} = 1$.
The perpendicular distance $p$ from the center $(-1, -1)$ to the common chord $(2-\alpha)x - y - 1 = 0$ is:
$p = \frac{|(2-\alpha)(-1) - (-1) - 1|}{\sqrt{(2-\alpha)^2 + (-1)^2}} = \frac{|\alpha-2+1-1|}{\sqrt{(2-\alpha)^2+1}} = \frac{|\alpha-2|}{\sqrt{\alpha^2-4\alpha+5}}$.
The length of the common chord is $2\sqrt{r^2-p^2} = \frac{2}{\sqrt{5}}$.
Thus,$\sqrt{r^2-p^2} = \frac{1}{\sqrt{5}} \Rightarrow r^2-p^2 = \frac{1}{5}$.
Since $r=1$,we have $1 - p^2 = \frac{1}{5} \Rightarrow p^2 = \frac{4}{5}$.
$\frac{(\alpha-2)^2}{\alpha^2-4\alpha+5} = \frac{4}{5} \Rightarrow 5(\alpha^2-4\alpha+4) = 4(\alpha^2-4\alpha+5)$.
$5\alpha^2-20\alpha+20 = 4\alpha^2-16\alpha+20$.
$\alpha^2 - 4\alpha = 0 \Rightarrow \alpha(\alpha-4) = 0$.
Since $\alpha \neq 0$,we get $\alpha = 4$.

(A) Given circles are:
$S_1: x^2+y^2-6x-4y+13-c^2=0$
$S_2: x^2+y^2-4x-6y+13-c^2=0$
Centres are $C_1(3,2)$ and $C_2(2,3)$ with radius $r_1=r_2=c$.
The equation of the common chord is given by $S_1-S_2=0$:
$(x^2+y^2-6x-4y+13-c^2)-(x^2+y^2-4x-6y+13-c^2)=0$
$-2x+2y=0 \Rightarrow x-y=0$.
Let $M$ be the midpoint of the common chord $AB$. Then $C_1M$ is the perpendicular distance from $C_1(3,2)$ to the line $x-y=0$:
$C_1M = \frac{|3-2|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.
In $\triangle AC_1M$,$AM^2 = AC_1^2 - C_1M^2 = c^2 - (\frac{1}{\sqrt{2}})^2 = c^2 - \frac{1}{2}$.
$AM = \sqrt{c^2 - \frac{1}{2}} = \sqrt{\frac{2c^2-1}{2}} = \frac{\sqrt{4c^2-2}}{2}$.
The length of the common chord $AB = 2AM = 2 \times \frac{\sqrt{4c^2-2}}{2} = \sqrt{4c^2-2}$.

(B) The equations of the circles are:
$S_1: x^2+y^2-2ax=0$
$S_2: x^2+y^2-2by=0$
The equation of the common chord is given by $S_1 - S_2 = 0$:
$(x^2+y^2-2ax) - (x^2+y^2-2by) = 0$
$-2ax + 2by = 0$
$ax - by = 0$
The centre of circle $S_1$ is $C_1(a, 0)$ and its radius is $r_1 = a$.
The perpendicular distance $d$ from the centre $C_1(a, 0)$ to the common chord $ax - by = 0$ is:
$d = \frac{|a(a) - b(0)|}{\sqrt{a^2 + (-b)^2}} = \frac{a^2}{\sqrt{a^2+b^2}}$
The length of the common chord is $2\sqrt{r_1^2 - d^2}$:
$= 2\sqrt{a^2 - \left(\frac{a^2}{\sqrt{a^2+b^2}}\right)^2}$
$= 2\sqrt{a^2 - \frac{a^4}{a^2+b^2}}$
$= 2\sqrt{\frac{a^2(a^2+b^2) - a^4}{a^2+b^2}}$
$= 2\sqrt{\frac{a^4 + a^2b^2 - a^4}{a^2+b^2}}$
$= 2\sqrt{\frac{a^2b^2}{a^2+b^2}}$
$= \frac{2ab}{\sqrt{a^2+b^2}}$

(D) The given equations of the circles are:
$x^2+y^2-4y=0$ $(1)$
$x^2+y^2-8x-4y+11=0$ $(2)$
Subtracting equation $(1)$ from $(2)$,we get the equation of the common chord:
$(x^2+y^2-8x-4y+11) - (x^2+y^2-4y) = 0$
$-8x+11=0$ $\Rightarrow 8x=11$ $\Rightarrow x=\frac{11}{8}$
For the first circle $x^2+y^2-4y=0$,the center is $C(0, 2)$ and the radius $r = \sqrt{0^2 + 2^2 - 0} = 2$.
The perpendicular distance $d$ from the center $(0, 2)$ to the line $x = \frac{11}{8}$ is:
$d = |0 - \frac{11}{8}| = \frac{11}{8}$
Let $L$ be the length of the common chord. The formula for the length of the chord is $L = 2\sqrt{r^2 - d^2}$.
$L = 2\sqrt{2^2 - (\frac{11}{8})^2} = 2\sqrt{4 - \frac{121}{64}}$
$L = 2\sqrt{\frac{256 - 121}{64}} = 2\sqrt{\frac{135}{64}} = 2 \times \frac{\sqrt{135}}{8} = \frac{\sqrt{135}}{4} \text{ units}$.

(C) Let the radii of the two circles be $r_1 = 15$ and $r_2 = 20$. The distance between their centers $C_1$ and $C_2$ is $d = 25$.
Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = d^2$.
Since the sum of the squares of the radii equals the square of the distance between the centers,the triangle $\triangle AC_1C_2$ is a right-angled triangle with $\angle C_1AC_2 = 90^\circ$.
The common chord $AB$ is perpendicular to the line joining the centers $C_1C_2$. Let the intersection point be $D$.
In $\triangle AC_1C_2$,the altitude $AD$ to the hypotenuse $C_1C_2$ represents half the length of the common chord.
Using the area of the triangle: $\text{Area} = \frac{1}{2} \times r_1 \times r_2 = \frac{1}{2} \times d \times AD$.
$15 \times 20 = 25 \times AD$.
$AD = \frac{300}{25} = 12$.
The length of the common chord is $2 \times AD = 2 \times 12 = 24$ units.

(A) Let the equation of the variable circle $S=0$ passing through the origin $(0,0)$ be $x^2+y^2+2gx+2fy=0$.
Since the circle touches the line $x-y=0$,the perpendicular distance from the center $(-g, -f)$ to the line must equal the radius $\sqrt{g^2+f^2}$.
$\frac{|-g+f|}{\sqrt{1^2+(-1)^2}} = \sqrt{g^2+f^2} \Rightarrow \frac{(g-f)^2}{2} = g^2+f^2$.
$g^2+f^2-2gf = 2g^2+2f^2$ $\Rightarrow g^2+f^2+2gf = 0$ $\Rightarrow (g+f)^2 = 0$ $\Rightarrow f = -g$.
Substituting $f = -g$ into the circle equation,we get $S = x^2+y^2+2gx-2gy = 0$.
The common chord of $S=0$ and $x^2+y^2+6x+8y-7=0$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2gx-2gy) - (x^2+y^2+6x+8y-7) = 0$.
$(2g-6)x - (2g+8)y + 7 = 0$.
$2g(x-y) - (6x+8y-7) = 0$.
This represents a family of lines passing through the intersection of $x-y=0$ and $6x+8y-7=0$.
Solving these: $x=y$,so $6x+8x-7=0$ $\Rightarrow 14x=7$ $\Rightarrow x=\frac{1}{2}$.
Thus,the fixed point is $\left(\frac{1}{2}, \frac{1}{2}\right)$.

(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$.
Since the circles intersect at $A$ and $B$,the equation of the common chord $AB$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$
$-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$.
The family of circles passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_1 - S_2) = 0$.
However,the circle with $AB$ as diameter is given by $S_1 + k(S_1 - S_2) = 0$ where the coefficient of $x^2$ and $y^2$ is $1$.
Alternatively,using the property that the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$,we substitute $x = -\frac{1}{2}$ into $S_1$:
$S_1 = (-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = y^2 + 3y + \frac{1}{4} = 0$.
The equation of the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$.
Substituting $x = -\frac{1}{2}$ into $S_1$ and $S_2$ gives the points $A$ and $B$.
The circle with diameter $AB$ is $x^2+y^2+2x+3y+1 + k(2x+1) = 0$.
Since the center lies on the line $x = -\frac{1}{2}$,we find $k$ such that the equation simplifies to $x^2+y^2+x+3y+\frac{1}{2} = 0$,which is $2x^2+2y^2+2x+6y+1 = 0$.

(C) The radical axis of two circles bisects all common tangents of the circles.
First,find the midpoint $M$ of $A(1, 1)$ and $B(0, 1)$:
$M = \left(\frac{1+0}{2}, \frac{1+1}{2}\right) = \left(\frac{1}{2}, 1\right)$.
Next,find the midpoint $N$ of $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$:
$N = \left(\frac{3/5 - 1/5}{2}, \frac{4/5 + 7/5}{2}\right) = \left(\frac{2/5}{2}, \frac{11/5}{2}\right) = \left(\frac{1}{5}, \frac{11}{10}\right)$.
The radical axis passes through $M$ and $N$. The slope $m$ of the line $MN$ is:
$m = \frac{11/10 - 1}{1/5 - 1/2} = \frac{1/10}{-3/10} = -\frac{1}{3}$.
The equation of the line $MN$ is:
$y - 1 = -\frac{1}{3}\left(x - \frac{1}{2}\right)$
$3(y - 1) = -x + \frac{1}{2}$
$3y - 3 = -x + 0.5$
$x + 3y = 3.5$
$2x + 6y = 7$.

(A) The given circle is $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=1, c=-7$.
The centre is $C_1(-g, -f) = (1, -1)$ and the radius is $r_1 = \sqrt{g^2+f^2-c} = \sqrt{1+1+7} = 3$.
The pair of lines $5x^2-xy-5x+y=0$ are normals to the circle $S=0$.
Factoring the equation: $x(5x-y) - 1(5x-y) = 0 \Rightarrow (x-1)(5x-y) = 0$.
The lines are $x=1$ and $y=5x$.
The intersection of these normals is the centre $C_2$ of circle $S=0$.
Solving $x=1$ and $y=5x$,we get $x=1, y=5$. So,$C_2(1, 5)$.
Since the circles touch externally,the distance between centres $C_1C_2 = r_1 + r_2$.
$C_1C_2 = \sqrt{(1-1)^2 + (5 - (-1))^2} = \sqrt{0^2 + 6^2} = 6$.
Thus,$6 = 3 + r_2 \Rightarrow r_2 = 3$.
The equation of circle $S=0$ with centre $(1, 5)$ and radius $3$ is $(x-1)^2 + (y-5)^2 = 3^2$.
$x^2 - 2x + 1 + y^2 - 10y + 25 = 9 \Rightarrow x^2 + y^2 - 2x - 10y + 17 = 0$.
The chord of contact of point $C_1(1, -1)$ with respect to $S=0$ is given by $T=0$.
$x(1) + y(-1) - 1(x+1) - 5(y-1) + 17 = 0$.
$x - y - x - 1 - 5y + 5 + 17 = 0$.
$-6y + 21 = 0$ $\Rightarrow 6y = 21$ $\Rightarrow 2y = 7$ $\Rightarrow 2y-7=0$.

(C) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ are the same.
Given $S_1 = px^2 + py^2 + 2g'x + 2f'y + d = 0$,we first normalize it by dividing by $p$:
$\frac{S_1}{p} = x^2 + y^2 + \frac{2g'}{p}x + \frac{2f'}{p}y + \frac{d}{p} = 0$.
Now,the equation of the common chord is $\frac{S_1}{p} - S_2 = 0$.
Multiplying by $p$,we get $S_1 - pS_2 = 0$.

(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.
Simplifying the second equation,we get $x^{2}+y^{2}=16$.
The equation of the common chord is obtained by subtracting the two equations: $(x^{2}+y^{2}-4x-4y) - (x^{2}+y^{2}-16) = 0$.
This simplifies to $-4x-4y+16=0$,or $x+y=4$.
Let the origin be $O(0,0)$. The circle $x^{2}+y^{2}-4x-4y=0$ passes through the origin. The common chord $x+y=4$ intersects the circle at points $A$ and $B$.
Since the circle $x^{2}+y^{2}-4x-4y=0$ can be written as $(x-2)^{2}+(y-2)^{2}=8$,its center is $C(2,2)$ and radius is $r = \sqrt{8} = 2\sqrt{2}$.
The distance from the center $C(2,2)$ to the line $x+y-4=0$ is $d = \frac{|2+2-4|}{\sqrt{1^{2}+1^{2}}} = 0$.
Since the distance is $0$,the common chord is a diameter of the first circle.
Any diameter subtends a right angle at any point on the circumference. Since the origin $(0,0)$ lies on the circle,the angle subtended by the diameter at the origin is $\frac{\pi}{2}$.

(C) The given equations of the circles are:
$C_{1}: x^{2}+y^{2}=4$ (centre $C_{1} = (0,0)$,radius $r_{1} = 2$)
$C_{2}: (x-2)^{2}+y^{2}=1$ (centre $C_{2} = (2,0)$,radius $r_{2} = 1$)
Let $N$ be the point of intersection of the common chord $PQ$ with the line joining the centres $C_{1}C_{2}$.
Subtracting the two equations:
$(x^{2}+y^{2}) - ((x-2)^{2}+y^{2}) = 4 - 1$
$x^{2} - (x^{2}-4x+4) = 3$
$4x - 4 = 3 \implies 4x = 7 \implies x = \frac{7}{4}$
Since $N$ lies on the line $C_{1}C_{2}$ (the $x$-axis),the coordinate of $N$ is $(\frac{7}{4}, 0)$.
The distance $C_{1}N = \frac{7}{4}$.
The distance $C_{2}N = |C_{1}C_{2} - C_{1}N| = |2 - \frac{7}{4}| = \frac{1}{4}$.
Both triangles $\Delta C_{1}PQ$ and $\Delta C_{2}PQ$ share the same base $PQ$.
The ratio of their areas is the ratio of their heights from the base $PQ$ to the vertices $C_{1}$ and $C_{2}$ respectively,which is the ratio of the distances $C_{1}N$ and $C_{2}N$:
$\frac{\text{Area}(\Delta C_{1}PQ)}{\text{Area}(\Delta C_{2}PQ)} = \frac{\frac{1}{2} \times PQ \times C_{1}N}{\frac{1}{2} \times PQ \times C_{2}N} = \frac{C_{1}N}{C_{2}N} = \frac{7/4}{1/4} = 7: 1$.
Thus,the correct option is $C$.