The equation of a line inclined at an angle f | vedclass

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(A) The slope of the line is $m = 	an(\frac{\pi}{4}) = 1$. Let the equation of the line be $y = x + c$,or $x - y + c = 0$.For the first circle $x^2 +

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$2x - 2y - 3 = 0$

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(A) The slope of the line is $m = 	an(\frac{\pi}{4}) = 1$. Let the equation of the line be $y = x + c$,or $x - y + c = 0$.For the first circle $x^2 + y^2 = 4$,the center is $O(0, 0)$ and radius $r_1 = 2$. The length of the intercept is $2\sqrt{r_1^2 - d_1^2}$,where $d_1$ is the perpendicular distance from $(0, 0)$ to $x - y + c = 0$.$d_1 = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$.Length of intercept $L_1 = 2\sqrt{4 - \frac{c^2}{2}}$.For the second circle $x^2 + y^2 - 10x - 14y + 65 = 0$,the center is $C(5, 7)$ and radius $r_2 = \sqrt{5^2 + 7^2 - 65} = \sqrt{25 + 49 - 65} = \sqrt{9} = 3$. The perpendicular distance $d_2$ from $(5, 7)$ to $x - y + c = 0$ is:$d_2 = \frac{|5 - 7 + c|}{\sqrt{2}} = \frac{|c - 2|}{\sqrt{2}}$.Length of intercept $L_2 = 2\sqrt{9 - \frac{(c - 2)^2}{2}}$.Since $L_1 = L_2$,we have $4 - \frac{c^2}{2} = 9 - \frac{(c - 2)^2}{2}$.$4 - \frac{c^2}{2} = 9 - \frac{c^2 - 4c + 4}{2} = 9 - \frac{c^2}{2} + 2c - 2$.$4 = 7 + 2c$.$2c = -3 \Rightarrow c = -\frac{3}{2}$.The equation of the line is $x - y - \frac{3}{2} = 0$,which simplifies to $2x - 2y - 3 = 0$.

The equation of a line inclined at an angle $\frac{\pi}{4}$ to the $X$-axis,such that the two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x - 14y + 65 = 0$ intercept equal lengths on it,is

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