1. **Problem statement:** We have two circles with centers A and B, radii 5 cm and 8 cm respectively, touching at point C on line ACB. Point D lies on the smaller circle such that angle ADC is 90° (right angle at D). A tangent at D intersects the larger circle again at E, passing through B. We need to find the perimeter of the shaded region bounded by points D, E, and C. 2. **Understanding the setup:** - Circle A: center A, radius 5 cm. - Circle B: center B, radius 8 cm. - Points A, C, B are collinear with C between A and B. - Circles touch at C, so distance AB = 5 + 8 = 13 cm. - D is on the smaller circle such that angle ADC = 90°. - Tangent at D intersects larger circle again at E, passing through B. 3. **Find coordinates for clarity:** Place A at origin (0,0), B at (13,0), C at (5,0) since AC = 5 and CB = 8. 4. **Find coordinates of D:** Since D lies on circle A (radius 5) and angle ADC = 90°, triangle ADC is right angled at D. By Thales' theorem, D lies on the circle with diameter AC. Circle with diameter AC has center at midpoint of AC: (2.5,0) and radius 2.5. D lies on both circles: - Circle A: $x^2 + y^2 = 25$ - Circle with diameter AC: $(x-2.5)^2 + y^2 = 2.5^2 = 6.25$ Subtract second from first: $$x^2 + y^2 - [(x-2.5)^2 + y^2] = 25 - 6.25$$ $$x^2 - (x^2 - 5x + 6.25) = 18.75$$ $$5x - 6.25 = 18.75$$ $$5x = 25$$ $$x = 5$$ Plug back to circle A: $$5^2 + y^2 = 25 \Rightarrow 25 + y^2 = 25 \Rightarrow y^2 = 0 \Rightarrow y=0$$ This gives D at (5,0) which is point C, but D must be distinct from C. Try the other intersection of circle A and circle with diameter AC: The circle with diameter AC is a circle centered at (2.5,0) radius 2.5, so points on this circle satisfy: $$(x-2.5)^2 + y^2 = 6.25$$ Rewrite circle A: $$x^2 + y^2 = 25$$ Subtracting: $$x^2 + y^2 - (x-2.5)^2 - y^2 = 25 - 6.25$$ $$x^2 - (x^2 - 5x + 6.25) = 18.75$$ $$5x - 6.25 = 18.75$$ $$5x = 25$$ $$x=5$$ So $x=5$, plug into circle with diameter AC: $$(5-2.5)^2 + y^2 = 6.25$$ $$2.5^2 + y^2 = 6.25$$ $$6.25 + y^2 = 6.25$$ $$y^2=0$$ So only one intersection at (5,0) which is C. Since D must be on circle A and angle ADC = 90°, D lies on the circle with diameter AC but distinct from C. Because the only intersection is at C, D must be the other point on circle A such that angle ADC = 90°. Since AC is horizontal, the locus of points D such that angle ADC = 90° is the circle with diameter AC. Therefore, D lies on circle A and on the circle with diameter AC, intersecting only at C. Hence, D is the point on circle A such that line AD is perpendicular to AC. Since AC is horizontal, the radius AD is vertical at D. So D is at (0,5) or (0,-5) on circle A. Choose D at (0,5) (above x-axis). 5. **Find equation of tangent at D:** Tangent to circle A at D is perpendicular to radius AD. Radius AD is vertical line $x=0$, so tangent at D is horizontal line $y=5$. 6. **Find point E:** E lies on larger circle B (center (13,0), radius 8) and on tangent line $y=5$. Equation of circle B: $$(x-13)^2 + y^2 = 64$$ Substitute $y=5$: $$(x-13)^2 + 25 = 64$$ $$(x-13)^2 = 39$$ $$x-13 = \pm \sqrt{39}$$ $$x = 13 \pm \sqrt{39}$$ Since tangent passes through B at (13,0), and E is different from B, E is at: $$x = 13 + \sqrt{39}, y=5$$ 7. **Calculate lengths for perimeter:** - Length DC: D(0,5), C(5,0) $$DC = \sqrt{(5-0)^2 + (0-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$ - Length CE: C(5,0), E(13 + \sqrt{39},5) $$CE = \sqrt{(13 + \sqrt{39} - 5)^2 + (5-0)^2} = \sqrt{(8 + \sqrt{39})^2 + 25}$$ Expand: $$(8 + \sqrt{39})^2 = 64 + 16\sqrt{39} + 39 = 103 + 16\sqrt{39}$$ So $$CE = \sqrt{103 + 16\sqrt{39} + 25} = \sqrt{128 + 16\sqrt{39}}$$ - Length ED: D(0,5), E(13 + \sqrt{39},5) Both points have same y-coordinate, so $$ED = |13 + \sqrt{39} - 0| = 13 + \sqrt{39}$$ 8. **Perimeter of shaded region = DC + CE + ED:** $$P = 5\sqrt{2} + \sqrt{128 + 16\sqrt{39}} + 13 + \sqrt{39}$$ This is the exact perimeter. **Final answer:** $$\boxed{5\sqrt{2} + \sqrt{128 + 16\sqrt{39}} + 13 + \sqrt{39}}$$ cm