If the angle between the circles x^2+y^2-2x-4 | vedclass

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(D) The given circles are $C_1: x^2+y^2-2x-4y+c=0$ and $C_2: x^2+y^2-4x-2y+4=0$. For $C_1$,the center is $O_1(1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c

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$\frac{7 \pm \sqrt{5}}{2}$

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(D) The given circles are $C_1: x^2+y^2-2x-4y+c=0$ and $C_2: x^2+y^2-4x-2y+4=0$. For $C_1$,the center is $O_1(1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$. For $C_2$,the center is $O_2(2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$. The distance between centers $d^2 = (2-1)^2 + (1-2)^2 = 1+1 = 2$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$. Given $	heta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \frac{(5-c) + 1 - 2}{2 \sqrt{5-c} \cdot 1} = \frac{4-c}{2 \sqrt{5-c}}$. Thus,$\sqrt{5-c} = 4-c$. Squaring both sides: $5-c = (4-c)^2 = 16 - 8c + c^2$. Rearranging gives $c^2 - 7c + 11 = 0$. Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{7 \pm \sqrt{49-44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60^{\circ}$,then $c$ is equal to

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