Question

The angle of intersection of the two circles \(x^2+y^2-2x-2y=0\) and \(x^2+y^2=4\) is:

• A. \(30^\circ\)
• B. \(60^\circ\)
• C. \(90^\circ\)
• D. \(45^\circ\)

Hint:

Orthogonal circles intersect at right angles.

Answer 1

The Correct Option is C

Step 1: Angle of intersection \(\theta\) satisfies: \[ \cos\theta=\frac{2(r_1^2+r_2^2-d^2)}{2r_1r_2} \]
Step 2: Substituting values gives \(\cos\theta=0\).
Step 3: \[ \theta=90^\circ \]