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(D) For the circle $x^2+y^2=4$,the center is $(0,0)$ and radius $r_1=2$. The perpendicular distance from the center $(0,0)$ to the line $2x-2y-3=0$ is

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$d_1=d_2$

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(D) For the circle $x^2+y^2=4$,the center is $(0,0)$ and radius $r_1=2$. The perpendicular distance from the center $(0,0)$ to the line $2x-2y-3=0$ is $p_1 = \frac{|2(0)-2(0)-3|}{\sqrt{2^2+(-2)^2}} = \frac{3}{\sqrt{8}}$.The length of the intercept $d_1 = 2\sqrt{r_1^2-p_1^2} = 2\sqrt{4-\frac{9}{8}} = 2\sqrt{\frac{23}{8}}$.For the circle $x^2+y^2-10x-14y+65=0$,the center is $(5,7)$ and radius $r_2 = \sqrt{5^2+7^2-65} = \sqrt{25+49-65} = \sqrt{9} = 3$.The perpendicular distance from the center $(5,7)$ to the line $2x-2y-3=0$ is $p_2 = \frac{|2(5)-2(7)-3|}{\sqrt{2^2+(-2)^2}} = \frac{|10-14-3|}{\sqrt{8}} = \frac{|-7|}{\sqrt{8}} = \frac{7}{\sqrt{8}}$.Wait,checking the calculation for $d_2$: $d_2 = 2\sqrt{r_2^2-p_2^2} = 2\sqrt{9-\frac{49}{8}} = 2\sqrt{\frac{72-49}{8}} = 2\sqrt{\frac{23}{8}}$.Since $d_1 = 2\sqrt{\frac{23}{8}}$ and $d_2 = 2\sqrt{\frac{23}{8}}$,we have $d_1=d_2$.

Suppose $d_1$ and $d_2$ are respectively the lengths of intercepts of the circles $x^2+y^2=4$ and $x^2+y^2-10x-14y+65=0$ on the line $2x-2y-3=0$. Then,which of the following is true?

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