Problem 52
Question
A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to \(9 .\) The locus of the point is a circle such that (A) centre of the circle coincides with that of square (B) centre of the circle is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) (C) radius of the circle is 2 (D) all the above are true
Step-by-Step Solution
Verified Answer
All options are true: the circle's center is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and its radius is 2.
1Step 1: Understand the Problem
We are given a square with side of unit length and a point such that the sum of the squares of its distances from the four sides is equal to 9. We need to find the characteristics of the locus of this point, which forms a circle.
2Step 2: Set up the Coordinate System
Place the square on the coordinate plane such that its vertices are at (0,0), (1,0), (1,1), and (0,1). This means the center of the square is at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
3Step 3: Derive Equation for Distances
Let the point be \((x, y)\) and calculate its distances to the four sides: \(y\), \(1-y\), \(x\), and \(1-x\).
4Step 4: Set Up the Locus Equation
The sum of the squares of the distances is \(y^2 + (1-y)^2 + x^2 + (1-x)^2 = 9\). Simplifying gives: \[y^2 + 1 - 2y + y^2 + x^2 + 1 - 2x + x^2 = 9\] \[2x^2 + 2y^2 - 2x - 2y + 2 = 9\].
5Step 5: Simplify to Standard Form of a Circle
Subtract 2 from both sides and divide by 2: \[x^2 + y^2 - x - y = \frac{7}{2}\] Completing the square for both \(x\) and \(y\) gives: \[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\].
6Step 6: Identify the Characteristics of the Circle
From the equation, \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\), we see the center of the circle is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and the radius is 2.
Key Concepts
Equation of a CircleCoordinate GeometryDistance Formula
Equation of a Circle
An equation of a circle in coordinate geometry represents the set of all points that are equidistant from a fixed point, called the center. The standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). Here, \((h, k)\) is the center of the circle, and \(r\) is the radius.
In the problem at hand, after simplifying the given conditions, we derived the equation \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\).
In the problem at hand, after simplifying the given conditions, we derived the equation \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\).
- Notice that the equation is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), with center \((\frac{1}{2}, \frac{1}{2})\).
- The right side of the equation is \(4\), which means our radius \(r\) is \(2\), since \(2^2 = 4\).
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves using a coordinate plane to explore geometric figures.
In coordinate geometry, each point is defined by an ordered pair \((x, y)\) in a plane, which allows us to study shapes and their properties analytically.
In coordinate geometry, each point is defined by an ordered pair \((x, y)\) in a plane, which allows us to study shapes and their properties analytically.
- For our problem, placing the square's vertices on the coordinate plane with coordinates \((0,0)\), \((1,0)\), \((1,1)\), and \((0,1)\) helps us visualize the problem.
- The center of the square being \((\frac{1}{2}, \frac{1}{2})\) aligns perfectly with the center of our derived circle equation.
Distance Formula
The distance formula in coordinate geometry is crucial for finding the distance between two points on a plane.
For points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
For points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
- This formula is particularly useful in this exercise to determine the distances from a moving point \((x, y)\) to the sides of the square.
- We measure vertical distances as \(y\) and \(1-y\) while horizontal distances are \(x\) and \(1-x\).
Other exercises in this chapter
Problem 50
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