Question

In the xy-plane, the length of the shortest path from (0, 0) to (12, 16) that does not go inside the circle ( x - 6) + ( y - 8) = 25 is:

• A. 5
• B. 10
• C. 5π
• D. 10 + 5π

Answer 1

Answer: (D)

10+5π

Answer 2

The problem asks for the shortest path from point A(0, 0) to point B(12, 16) that does not go inside the circle $(x - 6)^2 + (y - 8)^2 = 25$.

1. Identify the given points and circle:

   • Starting point: A = (0, 0)
   • Ending point: B = (12, 16)
   • Circle equation: $(x - 6)^2 + (y - 8)^2 = 25$
     • Center of the circle: C = (6, 8)
     • Radius of the circle: r = $\sqrt{25} = 5$

2. Check the positions of A and B relative to the circle:

   • Distance from A(0,0) to C(6,8): $d_{AC} = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. Since $d_{AC} = 10 > r = 5$, point A is outside the circle.
   • Distance from B(12,16) to C(6,8): $d_{BC} = \sqrt{(12-6)^2 + (16-8)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. Since $d_{BC} = 10 > r = 5$, point B is outside the circle.

3. Check if the straight line path AB intersects the circle:

   • The equation of the line passing through A(0,0) and B(12,16) is $y = \frac{16}{12}x = \frac{4}{3}x$, or $4x - 3y = 0$.
   • The distance from the center C(6,8) to the line $4x - 3y = 0$ is: $d = \frac{|4(6) - 3(8)|}{\sqrt{4^2 + (-3)^2}} = \frac{|24 - 24|}{\sqrt{16 + 9}} = \frac{0}{\sqrt{25}} = 0$.
   • Since the distance from the center to the line is 0, the line segment AB passes through the center of the circle C(6,8).

4. Determine the points of intersection and the path:

   • Since the line AB passes through the center C, the points of intersection of the line with the circle are $C \pm r \cdot \hat{u}_{AB}$, where $\hat{u}_{AB}$ is the unit vector along AB.

   • The vector $\vec{AB} = (12, 16)$. Its magnitude is $\sqrt{12^2 + 16^2} = \sqrt{144+256} = \sqrt{400} = 20$.

   • The unit vector $\hat{u}_{AB} = \left(\frac{12}{20}, \frac{16}{20}\right) = \left(\frac{3}{5}, \frac{4}{5}\right)$.

   • The intersection points are:

     • $P_1 = (6, 8) - 5\left(\frac{3}{5}, \frac{4}{5}\right) = (6, 8) - (3, 4) = (3, 4)$.
     • $P_2 = (6, 8) + 5\left(\frac{3}{5}, \frac{4}{5}\right) = (6, 8) + (3, 4) = (9, 12)$.

   • The straight line path from A to B is $A \to P_1 \to P_2 \to B$.

     • Length of $AP_1 = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
     • Length of $P_2B = \sqrt{(12-9)^2 + (16-12)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
     • The segment $P_1P_2$ has length $2r = 2 \times 5 = 10$. This segment is inside the circle.

   • To avoid going inside the circle, the path must follow an arc along the circle from $P_1$ to $P_2$.

   • The vectors from the center C to $P_1$ and $P_2$ are:

     • $\vec{CP_1} = (3-6, 4-8) = (-3, -4)$.
     • $\vec{CP_2} = (9-6, 12-8) = (3, 4)$.

   • Since $\vec{CP_2} = -\vec{CP_1}$, the points $P_1$ and $P_2$ are diametrically opposite on the circle.

   • The angle subtended by the arc $P_1P_2$ at the center is $\pi$ radians (180 degrees).

   • The length of this arc is $r \theta = 5 \times \pi = 5\pi$.

5. Calculate the total shortest path length:

   The shortest path consists of three segments:

   1. Straight line from A to $P_1$: Length $AP_1 = 5$.
   2. Arc along the circle from $P_1$ to $P_2$: Length $5\pi$.
   3. Straight line from $P_2$ to B: Length $P_2B = 5$.

   Total shortest path length = $AP_1 + \text{Arc length} + P_2B = 5 + 5\pi + 5 = 10 + 5\pi$.

The problem setup simplifies significantly because A, C, B are collinear, and A and B are symmetric with respect to C at a distance of $2r$.

The final answer is $10+5\pi$.