A circular table is pushed into a corner of the room, where two walls meet at a right angle. A point P on the edge of the table (as shown below) has a distance of 8 from one wall, and a distance of 9 from the other wall. Find the radius of the table.

1 answer

Let the radius of the circular table be denoted as \( r \). We will analyze the given information by placing the table's center at point \( O \) in a coordinate system where the two walls meet at the origin \( (0, 0) \).

Assuming one wall is aligned with the \( x \)-axis and the other wall is aligned with the \( y \)-axis, we can denote the distances from point \( P \) to the walls. The distance from point \( P \) to the wall along the \( x \)-axis (the horizontal wall) is given as 8, and the distance from point \( P \) to the wall along the \( y \)-axis (the vertical wall) is given as 9.

The coordinates of point \( P \) can hence be expressed as: \[ P = (r - 8, r - 9), \] where \( r \) is the radius of the circular table.

For point \( P \) to be located on the edge of the circular table, the following relationship must hold: \[ OP = r, \] where \( O \) (the center of the table) can be expressed as: \[ O = (r, r). \] The distance \( OP \) can then be calculated using the distance formula: \[ OP = \sqrt{(P_x - O_x)^2 + (P_y - O_y)^2}. \] Substituting the coordinates: \[ P_x = r - 8, \quad P_y = r - 9, \quad O_x = r, \quad O_y = r, \] we get \[ OP = \sqrt{((r - 8) - r)^2 + ((r - 9) - r)^2}. \] This simplifies to: \[ OP = \sqrt{(-8)^2 + (-9)^2} = \sqrt{64 + 81} = \sqrt{145}. \] Since we know \( OP = r \), we can equate the two: \[ \sqrt{145} = r. \]

Thus, the radius of the table is: \[ \boxed{\sqrt{145}}. \]