Question

Consider two circles S₁ and S₂ (externally touching) having centres at points A and B whose radii are 1 and 2 respectively. A tangent to circle S₁ from point B intersects the circle S₁ at point C. D is chosen on circle S₂ so that AC is parallel to BD and two segments BC and AD do not intersect. Segment AD intersect the circle S₁ at E. The line through B and E intersects the circle S₁ at another point F.

• A. The length of segment EF is $\frac{2\sqrt{3}}{3}$
• B. The area of triangle ABD is 2√2
• C. The length of the segment DE is 2
• D. ABD is a triangle of perimeter 2√3

Answer 1

Answer: (A), (B), (C)

(A), (B), (C)

Answer 2

Let $A$ be the center of circle $S_1$ with radius $r_1=1$, and $B$ be the center of circle $S_2$ with radius $r_2=2$. Since the circles touch externally, the distance between their centers is $AB = r_1 + r_2 = 1+2=3$. We place $A$ at the origin $(0,0)$ and $B$ on the positive x-axis at $(3,0)$. The equation of $S_1$ is $x^2+y^2=1$. The equation of $S_2$ is $(x-3)^2+y^2=4$.

$BC$ is a tangent to $S_1$ at $C$. Thus, $\triangle ABC$ is a right-angled triangle with $\angle ACB = 90^\circ$. Using Pythagoras theorem in $\triangle ABC$: $AB^2 = AC^2 + BC^2 \implies 3^2 = 1^2 + BC^2 \implies BC^2 = 8 \implies BC = 2\sqrt{2}$. Let $C=(x_C, y_C)$. Since $C$ is on $S_1$, $x_C^2+y_C^2=1$. The radius $AC$ is perpendicular to the tangent $BC$. Thus, $\vec{AC} \cdot \vec{BC} = 0$. $\vec{AC} = (x_C, y_C)$ and $\vec{BC} = (x_C-3, y_C)$. $x_C(x_C-3) + y_C^2 = 0 \implies x_C^2 - 3x_C + y_C^2 = 0$. Substituting $x_C^2+y_C^2=1$, we get $1 - 3x_C = 0$, so $x_C = 1/3$. Then $y_C^2 = 1 - (1/3)^2 = 8/9$, so $y_C = \pm \frac{2\sqrt{2}}{3}$. Let's choose $C = (1/3, \frac{2\sqrt{2}}{3})$. So, $\vec{AC} = (1/3, \frac{2\sqrt{2}}{3})$.

$AC$ is parallel to $BD$. Since $D$ is on $S_2$, $|\vec{BD}| = r_2 = 2$. $\vec{BD} = k \vec{AC}$ for some scalar $k$. $|\vec{BD}| = |k| |\vec{AC}| \implies 2 = |k| \cdot 1 \implies k = \pm 2$.

Case 1: $\vec{BD} = 2 \vec{AC}$. $\vec{D} = \vec{B} + 2\vec{AC} = (3,0) + 2(1/3, \frac{2\sqrt{2}}{3}) = (11/3, \frac{4\sqrt{2}}{3})$. Line $AD$ is $y = \frac{4\sqrt{2}}{11}x$. Line $BC$ is $y = -\frac{\sqrt{2}}{4}(x-3)$. These lines intersect for $x=11/9$, which lies on both segments $AD$ and $BC$. This case is rejected.

Case 2: $\vec{BD} = -2 \vec{AC}$. $\vec{D} = \vec{B} - 2\vec{AC} = (3,0) - 2(1/3, \frac{2\sqrt{2}}{3}) = (7/3, -\frac{4\sqrt{2}}{3})$. Line $AD$ is $y = -\frac{4\sqrt{2}}{7}x$. Line $BC$ is $y = -\frac{\sqrt{2}}{4}(x-3)$. These lines intersect for $x=-7/3$, which does not lie on segment $AD$ (as $x_D=7/3>0$) or segment $BC$. This case is accepted. Thus, $A=(0,0)$, $B=(3,0)$, $D=(7/3, -\frac{4\sqrt{2}}{3})$.

$E$ is the intersection of $AD$ and $S_1$. Line $AD$ is $y = -\frac{4\sqrt{2}}{7}x$. Substituting into $x^2+y^2=1$: $x^2 + (-\frac{4\sqrt{2}}{7}x)^2 = 1 \implies x^2 + \frac{32}{49}x^2 = 1 \implies \frac{81}{49}x^2 = 1 \implies x = \pm 7/9$. Since $E$ is on segment $AD$ starting from $A$, $x_E$ must have the same sign as $x_D$. So $x_E = 7/9$. $y_E = -\frac{4\sqrt{2}}{7} \cdot (7/9) = -\frac{4\sqrt{2}}{9}$. So $E=(7/9, -\frac{4\sqrt{2}}{9})$.

Line $BE$ intersects $S_1$ at $E$ and $F$. Line $BE$ passes through $B(3,0)$ and $E(7/9, -\frac{4\sqrt{2}}{9})$. Slope $m_{BE} = \frac{-4\sqrt{2}/9 - 0}{7/9 - 3} = \frac{-4\sqrt{2}/9}{-20/9} = \frac{\sqrt{2}}{5}$. Line $BE$: $y-0 = \frac{\sqrt{2}}{5}(x-3) \implies y = \frac{\sqrt{2}}{5}x - \frac{3\sqrt{2}}{5}$. Intersections with $x^2+y^2=1$: $x^2 + (\frac{\sqrt{2}}{5}x - \frac{3\sqrt{2}}{5})^2 = 1 \implies 27x^2 - 12x - 7 = 0$. One root is $x_E=7/9$. Let the other root be $x_F$. $x_E + x_F = 7/9 + x_F = 12/27 = 4/9 \implies x_F = 4/9 - 7/9 = -3/9 = -1/3$. $y_F = \frac{\sqrt{2}}{5}(-1/3 - 3) = \frac{\sqrt{2}}{5}(-10/3) = -\frac{2\sqrt{2}}{3}$. So $F=(-1/3, -\frac{2\sqrt{2}}{3})$.

\textbf{Evaluation of Options}

(A) Length of $EF$: $E=(7/9, -4\sqrt{2}/9)$, $F=(-1/3, -2\sqrt{2}/3)=(-3/9, -6\sqrt{2}/9)$. $EF^2 = (7/9 - (-3/9))^2 + (-4\sqrt{2}/9 - (-6\sqrt{2}/9))^2 = (10/9)^2 + (2\sqrt{2}/9)^2 = \frac{100}{81} + \frac{8}{81} = \frac{108}{81}$. $EF = \sqrt{\frac{108}{81}} = \frac{6\sqrt{3}}{9} = \frac{2\sqrt{3}}{3}$. Option (A) is TRUE.

(B) Area of $\triangle ABD$: $A=(0,0)$, $B=(3,0)$, $D=(7/3, -4\sqrt{2}/3)$. Area $= \frac{1}{2} |x_A(y_B-y_D) + x_B(y_D-y_A) + x_D(y_A-y_B)| = \frac{1}{2} |0 + 3(-4\sqrt{2}/3 - 0) + 0| = \frac{1}{2}|-4\sqrt{2}| = 2\sqrt{2}$. Option (B) is TRUE.

(C) Length of $DE$: $D=(7/3, -4\sqrt{2}/3)=(21/9, -12\sqrt{2}/9)$, $E=(7/9, -4\sqrt{2}/9)$. $DE^2 = (21/9 - 7/9)^2 + (-12\sqrt{2}/9 - (-4\sqrt{2}/9))^2 = (14/9)^2 + (-8\sqrt{2}/9)^2 = \frac{196}{81} + \frac{128}{81} = \frac{324}{81}=4$. $DE = \sqrt{4} = 2$. Option (C) is TRUE.

(D) Perimeter of $\triangle ABD$: $AB=3$, $BD=r_2=2$. $AD = |\vec{D}| = \sqrt{(7/3)^2 + (-4\sqrt{2}/3)^2} = \sqrt{49/9 + 32/9} = \sqrt{81/9} = 3$. Perimeter $= AB+BD+AD = 3+2+3=8$. $8 \neq 2\sqrt{3}$. Option (D) is FALSE.