Question

The acute angle between the curves \(x^2 + y^2 = x + y\) and \(x^2 + y^2 = 2y\) is:

• A. \(\frac{2\pi}{3}\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{4}\)

Hint:

When finding the angle between two circles, remember to complete the square to obtain their standard forms. The formula for the angle between two circles is based on the radii and the distance between their centers.

Answer 1

The Correct Option is D

We are given two curves, \(x^2 + y^2 = x + y\) and \(x^2 + y^2 = 2y\), and we need to find the acute angle between them. 
Step 1: Rewrite the equations of the curves in standard form. For the first curve: \[ x^2 + y^2 - x - y = 0. \] Complete the square to obtain: \[ (x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}. \] This represents a circle with center \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\). For the second curve: \[ x^2 + y^2 - 2y = 0. \] Complete the square to obtain: \[ x^2 + (y - 1)^2 = 1. \] This represents a circle with center \((0, 1)\) and radius \(1\). 
Step 2: Find the distance between the centers of the two circles. The center of the first circle is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and the center of the second circle is \((0, 1)\). The distance between the centers is: \[ d = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(\frac{1}{2} - 1\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2}. \] 
Step 3: Use the formula for the angle between two circles. The formula for the angle between two circles is: \[ \cos \theta = \frac{|r_1^2 + r_2^2 - d^2|}{2r_1r_2}, \] where \(r_1\) and \(r_2\) are the radii of the circles, and \(d\) is the distance between the centers. For the first circle, \(r_1 = \frac{1}{\sqrt{2}}\), and for the second circle, \(r_2 = 1\). Substituting the values into the formula: \[ \cos \theta = \frac{\left|\left(\frac{1}{\sqrt{2}}\right)^2 + 1^2 - \left(\frac{\sqrt{2}}{2}\right)^2\right|}{2 \times \frac{1}{\sqrt{2}} \times 1} = \frac{\left|\frac{1}{2} + 1 - \frac{1}{2}\right|}{\sqrt{2}} = \frac{1}{\sqrt{2}}. \] Thus, \[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}. \]