Math Problem Statement

Find the point on the circle (x - 5)^2 + (y - 4)^2 = 4 which is closest to the circle (x - 1)^2 + (y - 1)^2 = 1.

Solution

The given problem asks to find the point on the circle (x5)2+(y4)2=4(x - 5)^2 + (y - 4)^2 = 4 which is closest to the other circle (x1)2+(y1)2=1(x - 1)^2 + (y - 1)^2 = 1.

To solve this, follow these steps:

Step 1: Identify the centers and radii of the circles.

  • The first circle (x5)2+(y4)2=4(x - 5)^2 + (y - 4)^2 = 4 has center C1=(5,4)C_1 = (5, 4) and radius r1=2r_1 = 2 (since 4=2\sqrt{4} = 2).
  • The second circle (x1)2+(y1)2=1(x - 1)^2 + (y - 1)^2 = 1 has center C2=(1,1)C_2 = (1, 1) and radius r2=1r_2 = 1 (since 1=1\sqrt{1} = 1).

Step 2: Find the distance between the two centers.

The distance dd between the centers C1(5,4)C_1(5, 4) and C2(1,1)C_2(1, 1) can be found using the distance formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} Substitute the coordinates of C1C_1 and C2C_2: d=(51)2+(41)2=42+32=16+9=25=5d = \sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Step 3: Determine the closest point.

The closest point on the first circle to the second circle lies along the line between the centers. The direction vector from C1C_1 to C2C_2 is: v=(15,14)=(4,3)\vec{v} = (1 - 5, 1 - 4) = (-4, -3) Normalize this vector to get the unit direction vector: Magnitude of v=(4)2+(3)2=16+9=5\text{Magnitude of } \vec{v} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5 v^=(45,35)\hat{v} = \left( \frac{-4}{5}, \frac{-3}{5} \right) The closest point will be C1C_1 shifted by a distance of r1r_1 (radius of the first circle) in the direction of v^\hat{v}. Thus, the point is: P=C1+r1v^=(5,4)+2(45,35)P = C_1 + r_1 \cdot \hat{v} = (5, 4) + 2 \cdot \left( \frac{-4}{5}, \frac{-3}{5} \right) P=(5,4)+(85,65)=(5+85,4+65)P = (5, 4) + \left( \frac{-8}{5}, \frac{-6}{5} \right) = \left( 5 + \frac{-8}{5}, 4 + \frac{-6}{5} \right) P=(255+85,205+65)=(175,145)P = \left( \frac{25}{5} + \frac{-8}{5}, \frac{20}{5} + \frac{-6}{5} \right) = \left( \frac{17}{5}, \frac{14}{5} \right)

Thus, the point on the first circle closest to the second circle is (175,145)\left( \frac{17}{5}, \frac{14}{5} \right).

Let me know if you need more clarification or further details! Here are 5 related questions you may find interesting:

  1. How do you determine the point on a circle closest to a given point outside the circle?
  2. How does the distance between the centers of two circles relate to the radius of the circles when they are externally tangent?
  3. What is the geometric significance of the direction vector between two points in this problem?
  4. How can you extend this approach to find the closest point on a circle to a line instead of another circle?
  5. How do you solve problems involving tangents between two circles?

Tip: Always visualize the problem by sketching the circles, as it helps in understanding the relationship between the points and the circles.

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Math Problem Analysis

Mathematical Concepts

Geometry
Distance between points
Circle equation

Formulas

Distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Circle equation: (x - h)^2 + (y - k)^2 = r^2

Theorems

The closest point on a circle lies along the line joining the centers of the circles.

Suitable Grade Level

Grades 10-12