Pole and Polar Questions in English

(A) The equation of the polar of a point $(\alpha_i, \beta_i)$ with respect to the circle $x^2+y^2-2g_ix+c_i^2=0$ is given by $\alpha_ix + \beta_iy - g_i(x+\alpha_i) + c_i^2 = 0$.
Rearranging the terms,we get $(\alpha_i - g_i)x + \beta_iy + (c_i^2 - \alpha_ig_i) = 0$.
Comparing this with the given line $x - y = 0$,we have the ratios of coefficients equal:
$\frac{\alpha_i - g_i}{1} = \frac{\beta_i}{-1} = \frac{c_i^2 - \alpha_ig_i}{0}$.
From the third part,we get $c_i^2 - \alpha_ig_i = 0$,which implies $c_i^2 = \alpha_ig_i$.
From the first two parts,$\alpha_i - g_i = -\beta_i$,which implies $\alpha_i + \beta_i = g_i$.
Therefore,$\frac{\alpha_i + \beta_i}{g_i} = \frac{g_i}{g_i} = 1$.
Summing this for $i=1, 2, 3$,we get $\sum_{i=1}^3 \frac{\alpha_i + \beta_i}{g_i} = 1 + 1 + 1 = 3$.

(A) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $T=0$,which is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.
Given the circle $x^2+y^2+4x-8y-5=0$,we have $g=2, f=-4, c=-5$.
The polar of $(k, k+1)$ is:
$kx + (k+1)y + 2(x+k) - 4(y+k+1) - 5 = 0$
$kx + ky + y + 2x + 2k - 4y - 4k - 4 - 5 = 0$
$(k+2)x + (k-3)y - 2k - 9 = 0$
Rearranging the terms to isolate $k$:
$k(x+y-2) + (2x-3y-9) = 0$
For this to be true for all real values of $k$,both coefficients must be zero:
$x+y-2 = 0$ and $2x-3y-9 = 0$.
Solving these equations:
From the first,$y = 2-x$.
Substituting into the second: $2x - 3(2-x) - 9 = 0 \Rightarrow 2x - 6 + 3x - 9 = 0 \Rightarrow 5x = 15 \Rightarrow x = 3$.
Then $y = 2-3 = -1$.
The point is $(3, -1)$.

(D) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Given the circle $x^2+y^2-4x-6y+1=0$,we have $g=-2, f=-3, c=1$.
For the point $(2k, k-4)$,the polar equation is:
$x(2k) + y(k-4) - 2(x+2k) - 3(y+k-4) + 1 = 0$
$2kx + ky - 4y - 2x - 4k - 3y - 3k + 12 + 1 = 0$
$k(2x + y - 7) - 2x - 7y + 13 = 0$
For this line to pass through a fixed point regardless of $k$,the coefficients of $k$ and the constant term must both be zero:
$2x + y - 7 = 0$
$-2x - 7y + 13 = 0$
Adding these equations: $-6y + 6 = 0 \Rightarrow y = 1$.
Substituting $y=1$ into $2x + y - 7 = 0$: $2x + 1 - 7 = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = 3$.
Thus,the polar always passes through the point $(3, 1)$.

(A) Two lines $lx + my + n = 0$ and $l_1x + m_1y + n_1 = 0$ are conjugate with respect to the circle $x^2 + y^2 = r^2$ if the pole of the first line lies on the second line.
The pole of the line $lx + my + n = 0$ with respect to the circle $x^2 + y^2 = r^2$ is given by $(x_1, y_1) = (-\frac{lr^2}{n}, -\frac{mr^2}{n})$.
Since this point lies on the line $l_1x + m_1y + n_1 = 0$,we have:
$l_1(-\frac{lr^2}{n}) + m_1(-\frac{mr^2}{n}) + n_1 = 0$
$-l_1lr^2 - m_1mr^2 + n_1n = 0$
$nn_1 = r^2(ll_1 + mm_1)$.

(C) The equation of the polar of the point $(x_1, y_1) = (1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is given by $xx_1 + yy_1 - 2(x + x_1) - 3(y + y_1) + 9 = 0$.
Substituting $(1, 2)$:
$x(1) + y(2) - 2(x + 1) - 3(y + 2) + 9 = 0$
$x + 2y - 2x - 2 - 3y - 6 + 9 = 0$
$-x - y + 1 = 0 \Rightarrow x + y - 1 = 0$.
The inverse of a point $(x_1, y_1)$ is the foot of the perpendicular from the center of the circle to the polar line,but more generally,it is the point $P'$ on the line joining the center $C$ to $P$ such that $CP \cdot CP' = r^2$.
Alternatively,the inverse point $(\alpha, \beta)$ is the foot of the perpendicular from the point $(1, 2)$ to the polar line $x + y - 1 = 0$.
Using the formula $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = -\frac{(1 + 2 - 1)}{1^2 + 1^2} = -\frac{2}{2} = -1$.
$\alpha - 1 = -1 \Rightarrow \alpha = 0$.
$\beta - 2 = -1 \Rightarrow \beta = 1$.
Thus,the inverse point is $(0, 1)$.

(C) The inverse point of a point $P(x_1, y_1)$ with respect to a circle $S = 0$ is the point $P'$ such that $P'$ lies on the line joining the center of the circle to $P$,and the product of their distances from the center is equal to the square of the radius.
Alternatively,the inverse point is the foot of the perpendicular from the center of the circle to the polar of $P$ if $P$ is outside,or more generally,the point $P'$ such that the polar of $P'$ is the line passing through $P$ perpendicular to the line joining the center to $P$.
Given circle: $x^2 + y^2 - 4x - 6y + 9 = 0$. Center $C = (2, 3)$,Radius $r = \sqrt{2^2 + 3^2 - 9} = \sqrt{4 + 9 - 9} = 2$.
Polar of $P(1, 2)$ with respect to the circle is $x(1) + y(2) - 2(x + 1) - 3(y + 2) + 9 = 0$.
$x + 2y - 2x - 2 - 3y - 6 + 9 = 0$ $\Rightarrow -x - y + 1 = 0$ $\Rightarrow x + y - 1 = 0$.
The inverse point $P'(x', y')$ lies on the line passing through $C(2, 3)$ and $P(1, 2)$. The slope of $CP$ is $\frac{3-2}{2-1} = 1$. The equation of line $CP$ is $y - 2 = 1(x - 1) \Rightarrow y = x + 1$.
Intersection of $x + y - 1 = 0$ and $y = x + 1$: $x + (x + 1) - 1 = 0$ $\Rightarrow 2x = 0$ $\Rightarrow x = 0$. Then $y = 1$.
Thus,the inverse point is $(0, 1)$.

(B) Let the point on the circle $x^2+y^2=p^2$ be $(x_1, y_1)$.
Then $x_1^2+y_1^2=p^2$.
The equation of the polar of $(x_1, y_1)$ with respect to the circle $x^2+y^2=q^2$ is $x x_1+y y_1=q^2$.
This line touches the circle $x^2+y^2=r^2$.
The perpendicular distance from the center $(0, 0)$ to the line $x x_1+y y_1-q^2=0$ must be equal to the radius $r$.
So,$\frac{|0(x_1)+0(y_1)-q^2|}{\sqrt{x_1^2+y_1^2}} = r$.
$|q^2| = r \sqrt{x_1^2+y_1^2}$.
Since $x_1^2+y_1^2=p^2$,we have $q^2 = r \sqrt{p^2} = rp$.
Thus,$q^2 = pr$,which implies that $p, q, r$ are in $GP$.

(A) The equation of the line is $x+4y-4=0$. Comparing this with $lx+my+n=0$,we get $l=1, m=4, n=-4$.
The equation of the ellipse is $x^2+4y^2=4$,which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$.
Here,$a^2=4$ and $b^2=1$.
The formula for the pole $(x_1, y_1)$ of the line $lx+my+n=0$ with respect to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by:
$x_1 = -\frac{a^2l}{n}$ and $y_1 = -\frac{b^2m}{n}$.
Substituting the values:
$x_1 = -\frac{4 \times 1}{-4} = 1$
$y_1 = -\frac{1 \times 4}{-4} = 1$
Thus,the pole is $(1,1)$.

(C) The equation of the polar of a point $(\alpha, \beta)$ with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{\alpha x}{a^2} - \frac{\beta y}{b^2} = 1$.
Given hyperbola is $9x^2 - 16y^2 = 144$,which can be written as $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
So,the equation of the polar of $(\alpha, \beta)$ is $\frac{\alpha x}{16} - \frac{\beta y}{9} = 1$,or $9\alpha x - 16\beta y - 144 = 0$.
Comparing this with the given line $3x - 16y + 48 = 0$,we have:
$\frac{9\alpha}{3} = \frac{-16\beta}{-16} = \frac{-144}{48}$.
$3\alpha = \beta = -3$.
Thus,$\alpha = -1$ and $\beta = -3$.
Therefore,$\alpha - \beta = -1 - (-3) = -1 + 3 = 2$.

(C) The equation of the straight lines joining the origin to the points of intersection of the line $x \cos \alpha + y \sin \alpha = P$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is obtained by homogenizing the hyperbola equation using the line equation: $\frac{x^2}{9} - \frac{y^2}{36} = (\frac{x \cos \alpha + y \sin \alpha}{P})^2$.
Expanding this,we get $x^2(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + y^2(-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) - \frac{2xy \cos \alpha \sin \alpha}{P^2} = 0$.
Since the lines subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + (-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) = 0$.
This simplifies to $\frac{1}{12} - \frac{1}{P^2} = 0$,so $P^2 = 12$.
The polar of a point $(h, k)$ with respect to $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is $\frac{xh}{9} - \frac{yk}{36} = 1$.
Comparing this with $x \cos \alpha + y \sin \alpha = P$,we have $\frac{h/9}{\cos \alpha} = \frac{-k/36}{\sin \alpha} = \frac{1}{P}$.
Thus,$\cos \alpha = \frac{Ph}{9}$ and $\sin \alpha = -\frac{Pk}{36}$.
Using $\cos^2 \alpha + \sin^2 \alpha = 1$,we get $\frac{P^2 h^2}{81} + \frac{P^2 k^2}{1296} = 1$.
Substituting $P^2 = 12$,we get $\frac{12h^2}{81} + \frac{12k^2}{1296} = 1$,which simplifies to $\frac{4h^2}{27} + \frac{k^2}{108} = 1$.
Multiplying by $108$,we get $16h^2 + k^2 = 108$.
Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2 + y^2 = 108$.