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

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(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$. For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqr

Vedclass · Accepted Answer

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

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(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$. For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$. For $S_2$,the center $C_2 = (2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$. The distance between the centers $d = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. The angle $	heta$ between two circles is given by $\cos 	heta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} ight|$. Given $	heta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \left| \frac{(5-c) + 1 - 2}{2 \cdot \sqrt{5-c} \cdot 1} ight| = \left| \frac{4-c}{2\sqrt{5-c}} ight|$. Squaring both sides: $\frac{1}{4} = \frac{(4-c)^2}{4(5-c)} \Rightarrow 5-c = (4-c)^2$. $5-c = 16 - 8c + c^2 \Rightarrow c^2 - 7c + 11 = 0$. 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 $c=$

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