The shortest distance from a point to a line

Question:

P (x, y) is a point moving on the circle x² + y² = 1. What is the shortest distance from the point P to the line 3x - 4y - 12 = 0?

Solution:

Given: the circle equation x² + y² = 1, then its center is (0, 0) and radius is 1.

The given line equation is 3x - 4y - 12 = 0.

3x - 4y - 12 = 0

3x - 4y = 12

3x/12 - 4y/12 = 12/12

x/4 - y/3 = 1

x/4 + y/(-3) = 1

The line 3x - 4y - 12 = 0 intersects the x-axis at 4 and intersects the y-axis at -3.

Draw a line from the center of the circle and perpendicular to the line 3x - 4y - 12 = 0. This line intersects the circle at point E and perpendicular to the line at the point F.

When the point P move to the point E, the shortest distance from the point P to the line is |EF|.

Point O is the center of the circle. r is the radius of the circle.

|EF| = |OF| - |OE| = |OF| - r

|OF| is the distance from the point O (center of the circle) to the line 3x - 4y - 12 = 0.

d = |OF| = |Ax_o + By_o + C|/square root of (A² + B²⁾

From the circle equation x² + y² = 1, we get x_o = 0 and y_o = 0

From the line equation 3x - 4y - 12 = 0, we get A = 3, B = -4 and C = -12

d = |-12|/square root of (3² + (-4)²)

= 12/5

|EF| = |OF| - r

= 12/5 - 1

= 12/5 - 5/5

7/5 = = 1.4

Therefore, when the point P moves to the point E, the shortest distance from the point P to the line is 1.4. Watch the video for more details.