A rectangle is formed by the lines x=4, x=-2, | vedclass

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(B) The vertices of the rectangle are the intersection points of the given lines: $(4, 5), (4, -2), (-2, 5), (-2, -2)$.Since the rectangle is formed b

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$\left(1, \frac{-32}{7}\right)$

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(B) The vertices of the rectangle are the intersection points of the given lines: $(4, 5), (4, -2), (-2, 5), (-2, -2)$.Since the rectangle is formed by lines parallel to the axes,the circle passing through these vertices has the diagonal connecting $(-2, -2)$ and $(4, 5)$ as its diameter.The equation of the circle in diametrical form is $(x-4)(x+2) + (y-5)(y+2) = 0$.Expanding this,we get $x^2 - 2x - 8 + y^2 - 3y - 10 = 0$,which simplifies to $x^2 + y^2 - 2x - 3y - 18 = 0$.Let the pole of the line $y+2=0$ be $(h, k)$.The equation of the polar of a point $(h, k)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xh + yk + g(x+h) + f(y+k) + c = 0$.Substituting the values $g=-1, f=-1.5, c=-18$,we get $xh + yk - 1(x+h) - 1.5(y+k) - 18 = 0$.Rearranging,$(h-1)x + (k-1.5)y - (h + 1.5k + 18) = 0$.Comparing this with the given line $0x + 1y + 2 = 0$,we have $\frac{h-1}{0} = \frac{k-1.5}{1} = \frac{-(h + 1.5k + 18)}{2}$.From $\frac{h-1}{0}$,we get $h-1 = 0$,so $h=1$.Substituting $h=1$ into the second equality: $k-1.5 = \frac{-(1 + 1.5k + 18)}{2}$ $\Rightarrow 2k - 3 = -19 - 1.5k$ $\Rightarrow 3.5k = -16$ $\Rightarrow k = \frac{-16}{3.5} = \frac{-32}{7}$.Thus,the pole is $\left(1, \frac{-32}{7}\right)$.

$A$ rectangle is formed by the lines $x=4, x=-2, y=5$,and $y=-2$. $A$ circle is drawn passing through the vertices of this rectangle. The pole of the line $y+2=0$ with respect to this circle is:

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