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Question-223619

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Question Number 223619 by Tawa11 last updated on 01/Aug/25
Answered by Frix last updated on 01/Aug/25
∣C_1 C_2 ∣=(√2) ⇒ C_1 , C_2 form a triangle with each of the intersection points with a=b=(√2), c=2 ⇒ the angle is 45° The angle between the circles = the angle of the tangents at the intersection = the angle of the radius vectors at the intersection. circle_1 : C_1 , r_1 and circle_2 : C_2 , r_2 ⇒ the distances to the intersections I_1 , I_2 are (without calculation): ∣C_1 I_k ∣=r_1 ∣C_2 I_k ∣=r_2 and we needto calculate ∣C_1 C_2 ∣=d the angle between the circles is given by the cosine law r_1 ^2 +r_2 ^2 −2r_1 r_2 cos θ =d^2 θ=cos^(−1) ((r_1 ^2 +r_2 ^2 −d^2 )/(2r_1 r_2 ))
$$\mid{C}_{\mathrm{1}} {C}_{\mathrm{2}} \mid=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{C}_{\mathrm{1}} ,\:{C}_{\mathrm{2}} \:\mathrm{form}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{intersection}\:\mathrm{points}\:\mathrm{with} \\ $$$${a}={b}=\sqrt{\mathrm{2}},\:{c}=\mathrm{2}\:\Rightarrow\:\mathrm{the}\:\mathrm{angle}\:\mathrm{is}\:\mathrm{45}° \\ $$$$ \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circles}\:=\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the}\:\mathrm{intersection}\:=\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{vectors}\:\mathrm{at}\:\mathrm{the}\:\mathrm{intersection}. \\ $$$$\mathrm{circle}_{\mathrm{1}} :\:{C}_{\mathrm{1}} ,\:{r}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{circle}_{\mathrm{2}} :\:{C}_{\mathrm{2}} ,\:{r}_{\mathrm{2}} \:\Rightarrow\:\mathrm{the} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{intersections}\:{I}_{\mathrm{1}} ,\:{I}_{\mathrm{2}} \:\mathrm{are} \\ $$$$\left(\mathrm{without}\:\mathrm{calculation}\right): \\ $$$$\mid{C}_{\mathrm{1}} {I}_{{k}} \mid={r}_{\mathrm{1}} \\ $$$$\mid{C}_{\mathrm{2}} {I}_{{k}} \mid={r}_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{needto}\:\mathrm{calculate}\:\:\mid{C}_{\mathrm{1}} {C}_{\mathrm{2}} \mid={d} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circles}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{cosine}\:\mathrm{law} \\ $$$${r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{cos}\:\theta\:={d}^{\mathrm{2}} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \:\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}{r}_{\mathrm{1}} {r}_{\mathrm{2}} } \\ $$
Commented by Tawa11 last updated on 01/Aug/25
Wow, I understand now. Thanks sir.
$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by Raphael254 last updated on 01/Aug/25
The angle between two circles is the angle formed by the two tangents of one of the two points of intercept of the two circles. In our case, the angle is a sum of the angle of a square and a rectangle triangle: angle of square = 90° angle of rectangle triangle = 45° angle between the two circles = 135°
$${The}\:{angle}\:{between}\:{two}\:{circles}\:{is}\:{the}\:{angle}\:{formed}\:{by}\:{the}\:{two}\:{tangents}\:{of}\:{one}\:{of}\:{the}\:{two}\:{points}\:{of}\:{intercept}\:{of}\:{the}\:{two}\:{circles}. \\ $$$$ \\ $$$${In}\:{our}\:{case},\:{the}\:{angle}\:{is}\:{a}\:{sum}\:{of}\:{the}\:{angle}\:{of}\:{a}\:{square}\:{and}\:{a}\:{rectangle}\:{triangle}: \\ $$$$ \\ $$$${angle}\:{of}\:{square}\:=\:\mathrm{90}° \\ $$$${angle}\:{of}\:{rectangle}\:{triangle}\:=\:\mathrm{45}° \\ $$$$ \\ $$$$\:{angle}\:{between}\:{the}\:{two}\:{circles}\:=\:\mathrm{135}° \\ $$
Commented by Frix last updated on 01/Aug/25
The angle between two lines is “not unique” because there are two angles α, β with α+β=180° 45°+135°=180° ⇒ both answers are ok
$$\mathrm{The}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{is}\:“\mathrm{not}\:\mathrm{unique}'' \\ $$$$\mathrm{because}\:\mathrm{there}\:\mathrm{are}\:\mathrm{two}\:\mathrm{angles}\:\alpha,\:\beta\:\mathrm{with} \\ $$$$\alpha+\beta=\mathrm{180}° \\ $$$$\mathrm{45}°+\mathrm{135}°=\mathrm{180}°\:\Rightarrow\:\mathrm{both}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{ok} \\ $$
Commented by Raphael254 last updated on 01/Aug/25
Commented by Tawa11 last updated on 01/Aug/25
Wow. Thanks sir. I appreciate. But when I followed sir MJS part. I got 45°.
$$\mathrm{Wow}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$$$\mathrm{But}\:\mathrm{when}\:\mathrm{I}\:\mathrm{followed}\:\mathrm{sir}\:\mathrm{MJS}\:\mathrm{part}. \\ $$$$\mathrm{I}\:\mathrm{got}\:\:\:\mathrm{45}°. \\ $$
Commented by Tawa11 last updated on 01/Aug/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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