Question

In the diagram, DC is a diameter of the large circle centered at A, and AC is a diameter of the smaller circle centered at B. If DE is tangent to the smaller circle at F and DC = 12 then the length of DE is:

• A. 13
• B. 16
• C. 8√2
• D. 10√2

Answer 1

Answer: (C)

8√2

Answer 2

1. Establish Coordinates and Radii: Let the line containing the diameters DC and AC be the x-axis. Since DC is the diameter of the large circle centered at A, and DC = 12, the radius of the large circle is $R = 12/2 = 6$. A is the midpoint of DC. Let A be at the origin (0, 0). Then D is at (-6, 0) and C is at (6, 0). AC is the diameter of the smaller circle centered at B. The length of AC is $|6 - 0| = 6$. The radius of the smaller circle is $r = 6/2 = 3$. B is the midpoint of AC, so B is at $((0+6)/2, 0) = (3, 0)$. The large circle has center A(0, 0) and radius $R=6$. Its equation is $x^2 + y^2 = 36$. The smaller circle has center B(3, 0) and radius $r=3$. Its equation is $(x-3)^2 + y^2 = 9$. The point D is at (-6, 0).

2. Tangent to the Smaller Circle: DE is tangent to the smaller circle at F. This means the line segment DE is tangent to the circle $(x-3)^2 + y^2 = 9$. The distance from the center B(3, 0) to the point of tangency F is the radius $r=3$, and BF is perpendicular to the tangent line DE. Consider the right-angled triangle DFB, with the right angle at F. The distance DB is the distance between D(-6, 0) and B(3, 0), which is $DB = |3 - (-6)| = 9$. In right triangle DFB, by Pythagorean theorem, $DF^2 + BF^2 = DB^2$. $DF^2 + 3^2 = 9^2$ $DF^2 + 9 = 81$ $DF^2 = 72$ $DF = \sqrt{72} = 6\sqrt{2}$. This is the length of the tangent segment from D to the smaller circle.

3. Point E on the Larger Circle: The problem states that DE is the length of the segment, and the diagram shows E is on the larger circle. Thus, the line DE is a chord of the larger circle that is tangent to the smaller circle. Let the equation of the line DE passing through D(-6, 0) be $y = m(x+6)$. The distance from B(3, 0) to this line $mx - y + 6m = 0$ must be equal to the radius $r=3$. $\frac{|m(3) - 0 + 6m|}{\sqrt{m^2 + (-1)^2}} = 3$ $\frac{|9m|}{\sqrt{m^2 + 1}} = 3$ $9|m| = 3\sqrt{m^2 + 1}$ $3|m| = \sqrt{m^2 + 1}$ Squaring both sides: $9m^2 = m^2 + 1 \implies 8m^2 = 1 \implies m^2 = 1/8 \implies m = \pm \frac{1}{\sqrt{8}} = \pm \frac{\sqrt{2}}{4}$.

4. Find Coordinates of E: Let's take $m = \frac{\sqrt{2}}{4}$. The line DE is $y = \frac{\sqrt{2}}{4}(x+6)$. Point E is the intersection of this line with the large circle $x^2 + y^2 = 36$. Substitute y: $x^2 + \left(\frac{\sqrt{2}}{4}(x+6)\right)^2 = 36$ $x^2 + \frac{2}{16}(x+6)^2 = 36$ $x^2 + \frac{1}{8}(x^2 + 12x + 36) = 36$ $8x^2 + x^2 + 12x + 36 = 288$ $9x^2 + 12x - 252 = 0$ $3x^2 + 4x - 84 = 0$ Solving for x using the quadratic formula: $x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-84)}}{2(3)} = \frac{-4 \pm \sqrt{16 + 1008}}{6} = \frac{-4 \pm \sqrt{1024}}{6} = \frac{-4 \pm 32}{6}$. The possible values for x are $x_1 = \frac{-4 + 32}{6} = \frac{28}{6} = \frac{14}{3}$ and $x_2 = \frac{-4 - 32}{6} = \frac{-36}{6} = -6$. $x_2 = -6$ corresponds to point D. So, $x_E = 14/3$. Now find $y_E$: $y_E = \frac{\sqrt{2}}{4}(\frac{14}{3}+6) = \frac{\sqrt{2}}{4}(\frac{14+18}{3}) = \frac{\sqrt{2}}{4}(\frac{32}{3}) = \frac{8\sqrt{2}}{3}$. So, $E = (14/3, 8\sqrt{2}/3)$.

5. Calculate Length DE: The coordinates of D are (-6, 0) and E are (14/3, $8\sqrt{2}/3$). $DE^2 = (x_E - x_D)^2 + (y_E - y_D)^2$ $DE^2 = (\frac{14}{3} - (-6))^2 + (\frac{8\sqrt{2}}{3} - 0)^2$ $DE^2 = (\frac{14}{3} + \frac{18}{3})^2 + (\frac{8\sqrt{2}}{3})^2$ $DE^2 = (\frac{32}{3})^2 + \frac{64 \times 2}{9}$ $DE^2 = \frac{1024}{9} + \frac{128}{9} = \frac{1152}{9} = 128$. $DE = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$.