Question

Let \( x - y = 0 \) and \( x + y = 1 \) be two perpendicular diameters of a circle of radius \( R \). The circle will pass through the origin if \( R \) is equal to:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{\sqrt{3}} \)
• D. \( \frac{1}{3} \)

Hint:

The intersection of perpendicular diameters gives the center of the circle. The distance from the center to any point on the circle (including the origin, if it lies on the circle) is the radius.

Answer 1

The Correct Option is B

Step 1: Determine the circle's center by finding the intersection of its diameters.
\nSolving \( x - y = 0 \) and \( x + y = 1 \) yields the center \( C \left( \frac{1}{2}, \frac{1}{2} \right) \).
\n\n
Step 2: Utilize the fact that the circle goes through the origin \( (0, 0) \).
\nThe radius \( R \) is the distance from the center to the origin.\n\n
Step 3: Compute the distance.
\n\[\nR = \sqrt{\left( \frac{1}{2} - 0 \right)^2 + \left( \frac{1}{2} - 0 \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}\n\]