1. **Problem statement:** Given a circle with center M, a line passing through the origin O(0,0) tangent to the circle at B(3,4), and the segment OM connecting the origin and center M. 2. **Find the equation of the line on which M lies.** Given: The line OM has equation $y=\frac{1}{2}x$. 3. **Find the center M and the equation of the circle.** - Since M lies on $y=\frac{1}{2}x$, let $M=(m, \frac{m}{2})$. - The radius is the distance MB. - The line from O to B is tangent to the circle at B, so the radius MB is perpendicular to OB. 4. **Calculate the slope of OB:** $$\text{slope}_{OB} = \frac{4-0}{3-0} = \frac{4}{3}$$ 5. **Slope of radius MB (perpendicular to OB):** $$\text{slope}_{MB} = -\frac{1}{\text{slope}_{OB}} = -\frac{3}{4}$$ 6. **Equation of line MB passing through B(3,4):** $$y-4 = -\frac{3}{4}(x-3)$$ $$y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4}$$ 7. **Find intersection of line MB and line OM to get M:** Set $$\frac{1}{2}x = -\frac{3}{4}x + \frac{25}{4}$$ Multiply both sides by 4: $$2x = -3x + 25$$ $$2x + 3x = 25$$ $$5x = 25$$ $$x = 5$$ Then $$y = \frac{1}{2} \times 5 = \frac{5}{2} = 2.5$$ So $$M = (5, 2.5)$$ 8. **Radius length $r = |MB|$:** $$r = \sqrt{(5-3)^2 + (2.5-4)^2} = \sqrt{2^2 + (-1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5$$ 9. **Equation of the circle:** $$ (x-5)^2 + (y-2.5)^2 = (2.5)^2 = 6.25 $$ 10. **Find point C where line BM extended intersects the circle again:** - Parametric form of BM: from B(3,4) to M(5,2.5), vector is $(2, -1.5)$. - Let $C = B + t(2, -1.5) = (3+2t, 4 - 1.5t)$ with $t \neq 0$. - Substitute into circle equation: $$ (3+2t -5)^2 + (4 - 1.5t - 2.5)^2 = 6.25 $$ $$ (2t - 2)^2 + (1.5 - 1.5t)^2 = 6.25 $$ $$ (2t - 2)^2 + (1.5(1 - t))^2 = 6.25 $$ $$ (2t - 2)^2 + 2.25(1 - t)^2 = 6.25 $$ Expand: $$ (2t - 2)^2 = 4t^2 - 8t + 4 $$ $$ 2.25(1 - 2t + t^2) = 2.25 - 4.5t + 2.25t^2 $$ Sum: $$ 4t^2 - 8t + 4 + 2.25 - 4.5t + 2.25t^2 = 6.25 $$ $$ (4 + 2.25) t^2 + (-8 - 4.5) t + (4 + 2.25) = 6.25 $$ $$ 6.25 t^2 - 12.5 t + 6.25 = 6.25 $$ Subtract 6.25: $$ 6.25 t^2 - 12.5 t = 0 $$ Factor: $$ 6.25 t (t - 2) = 0 $$ Solutions: $$ t=0 \quad \text{or} \quad t=2 $$ Ignore $t=0$ (point B), so $t=2$. 11. **Coordinates of C:** $$ x = 3 + 2 \times 2 = 7 $$ $$ y = 4 - 1.5 \times 2 = 4 - 3 = 1 $$ So $$ C = (7, 1) $$ 12. **Find point D where line BC intersects x-axis:** - Slope of BC: $$ m_{BC} = \frac{1 - 4}{7 - 3} = \frac{-3}{4} = -\frac{3}{4} $$ - Equation of BC: $$ y - 4 = -\frac{3}{4}(x - 3) $$ - On x-axis, $y=0$: $$ 0 - 4 = -\frac{3}{4}(x - 3) $$ $$ -4 = -\frac{3}{4}x + \frac{9}{4} $$ Multiply both sides by 4: $$ -16 = -3x + 9 $$ $$ -16 - 9 = -3x $$ $$ -25 = -3x $$ $$ x = \frac{25}{3} $$ So $$ D = \left(\frac{25}{3}, 0\right) $$ 13. **Calculate areas of triangles OMD and ODC:** - Coordinates: $$ O = (0,0), M = (5, 2.5), D = \left(\frac{25}{3}, 0\right), C = (7,1) $$ - Area of triangle OMD: $$ S_{OMD} = \frac{1}{2} |x_O(y_M - y_D) + x_M(y_D - y_O) + x_D(y_O - y_M)| $$ $$ = \frac{1}{2} |0(2.5 - 0) + 5(0 - 0) + \frac{25}{3}(0 - 2.5)| $$ $$ = \frac{1}{2} |0 + 0 - \frac{25}{3} \times 2.5| = \frac{1}{2} \times \frac{62.5}{3} = \frac{62.5}{6} $$ - Area of triangle ODC: $$ S_{ODC} = \frac{1}{2} |0(0 - 1) + \frac{25}{3}(1 - 0) + 7(0 - 0)| $$ $$ = \frac{1}{2} |0 + \frac{25}{3} + 0| = \frac{1}{2} \times \frac{25}{3} = \frac{25}{6} $$ 14. **Ratio of areas:** $$ \frac{S_{OMD}}{S_{ODC}} = \frac{\frac{62.5}{6}}{\frac{25}{6}} = \frac{62.5}{25} = 2.5 $$ **Final answers:** - Line OM: $y=\frac{1}{2}x$ - Center $M = (5, 2.5)$ - Circle equation: $$(x-5)^2 + (y-2.5)^2 = 6.25$$ - Point $C = (7,1)$ - Ratio of areas $\frac{S_{OMD}}{S_{ODC}} = 2.5$