The angle between the circles x^2+y^2-2x-9=0 | vedclass

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(B) The given circles are $S_1: x^2+y^2-2x-9=0$ and $S_2: x^2+y^2-4y-1=0$.The centre $C_1$ and radius $r_1$ of $S_1$ are $(1, 0)$ and $\sqrt{1^2+0^2-(

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$\frac{\pi}{4}$

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(B) The given circles are $S_1: x^2+y^2-2x-9=0$ and $S_2: x^2+y^2-4y-1=0$.The centre $C_1$ and radius $r_1$ of $S_1$ are $(1, 0)$ and $\sqrt{1^2+0^2-(-9)} = \sqrt{10}$.The centre $C_2$ and radius $r_2$ of $S_2$ are $(0, 2)$ and $\sqrt{0^2+2^2-(-1)} = \sqrt{5}$.The distance $d$ between the centres $C_1(1, 0)$ and $C_2(0, 2)$ is $d = \sqrt{(1-0)^2 + (0-2)^2} = \sqrt{1+4} = \sqrt{5}$.The angle $	heta$ between the circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.Substituting the values: $\cos 	heta = \frac{10 + 5 - 5}{2 	imes \sqrt{10} 	imes \sqrt{5}} = \frac{10}{2 	imes \sqrt{50}} = \frac{10}{2 	imes 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \frac{\pi}{4}$.

The angle between the circles $x^2+y^2-2x-9=0$ and $x^2+y^2-4y-1=0$ at their point of intersection is

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