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(A) The lines $x+y=3$ and $x-y=3$ intersect at $(3, 0)$. The angle bisectors of these lines are $y=0$ and $x=3$. Since the circles touch both lines,th

Vedclass · Accepted Answer

$768$

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(A) The lines $x+y=3$ and $x-y=3$ intersect at $(3, 0)$. The angle bisectors of these lines are $y=0$ and $x=3$. Since the circles touch both lines,their centers must lie on the angle bisector $y=0$. Let the center be $(a, 0)$.The radius $r$ is the perpendicular distance from $(a, 0)$ to the line $x+y-3=0$,so $r = \frac{|a+0-3|}{\sqrt{1^2+1^2}} = \frac{|a-3|}{\sqrt{2}}$.The equation of the circle is $(x-a)^2 + y^2 = r^2 = \frac{(a-3)^2}{2}$.Since the circle passes through $(-9, 4)$,we have $(-9-a)^2 + 4^2 = \frac{(a-3)^2}{2}$.$2((a+9)^2 + 16) = (a-3)^2$$2(a^2 + 18a + 81 + 16) = a^2 - 6a + 9$$2a^2 + 36a + 194 = a^2 - 6a + 9$$a^2 + 42a + 185 = 0$$(a+37)(a+5) = 0$Thus,$a = -37$ or $a = -5$.For $a = -37$,$r_1 = \frac{|-37-3|}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2}$,so $r_1^2 = 800$.For $a = -5$,$r_2 = \frac{|-5-3|}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$,so $r_2^2 = 32$.The absolute difference between the squares of the radii is $|800 - 32| = 768$.

The absolute difference between the squares of the radii of the two circles passing through the point $(-9, 4)$ and touching the lines $x+y=3$ and $x-y=3$ is equal to . . . . . . .

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