Question
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.
Answers
GPT-4o mini
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:
\[
\boxed{\sqrt{145}}.
\]
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:
\[
\boxed{\sqrt{145}}.
\]