Question

A circle with center at \( (x, y) = (0.5, 0) \) and radius = 0.5 intersects with another circle with center at \( (x, y) = (1, 1) \) and radius = 1 at two points. One of the points of intersection \( (x, y) \) is:

• A. \( (0, 0) \)
• B. \( (0.2, 0.4) \)
• C. \( (0.5, 0.5) \)
• D. \( (1, 2) \)

Hint:

When solving for the intersection of two circles, expand the equations, eliminate terms, and solve the resulting system of linear equations.

Answer 1

The Correct Option is B

We are given two circles with the following equations: \[ (x - 0.5)^2 + y^2 = 0.5^2 \quad {(Equation 1: Circle 1)} \] \[ (x - 1)^2 + (y - 1)^2 = 1^2 \quad {(Equation 2: Circle 2)}. \] To solve this, we can expand both equations.
Expanding Equation 1: \[ (x - 0.5)^2 + y^2 = 0.25 + y^2 = 0.25 \quad \Rightarrow \quad x^2 - x + 0.25 + y^2 = 0.25 \quad \Rightarrow \quad x^2 - x + y^2 = 0. \] Expanding Equation 2: \( (x - 1)^2 + (y - 1)^2 = 1 \quad \Rightarrow \quad (x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 \quad \Rightarrow \quad x^2 - 2x + y^2 - 2y + 2 = 1 \quad \Rightarrow \quad x^2 - 2x + y^2 - 2y = -1. \) Now, subtract Equation 1 from Equation 2: \[ (x^2 - 2x + y^2 - 2y) - (x^2 - x + y^2) = -1 - 0 \quad \Rightarrow \quad -x - 2y = -1 \quad \Rightarrow \quad x + 2y = 1 \quad \cdots (3). \] Now, substitute \( x = 1 - 2y \) from Equation (3) into Equation 1: \[ (1 - 2y)^2 - (1 - 2y) + y^2 = 0. \] Expanding and solving for \( y \), we get: \[ 1 - 4y + 4y^2 - 1 + 2y + y^2 = 0 \quad \Rightarrow \quad 5y^2 - 2y = 0 \quad \Rightarrow \quad y(5y - 2) = 0. \] Thus, \( y = 0 \) or \( y = 0.4 \). For \( y = 0.4 \), substitute into \( x = 1 - 2y \) to get \( x = 0.2 \). Thus, the point of intersection is \( (0.2, 0.4) \).
The other point of intersection can be calculated similarly, but for this question, the correct answer is \( (0.2, 0.4) \).