The two circles intersect at two points P and Q. PR and PS are diameters of the two circles. What is ∠ PQR ?

Correct — C. Work inside the first circle alone, the one with diameter PR.

Q lies on that circle, since Q is one of the two points where the circles meet
PR is a diameter, so P and R are the ends of a semicircle
An angle standing on a diameter, from any other point of the circle, is a right angle

∠ PQR = 90° → option (c). The radius of either circle never enters the working.

Why the others are wrong

• (a)
  45° — 45° — you reach this only by splitting one right angle between the two circles. Each circle delivers its own full 90° at Q, independently of the other.
• (b)
  0° — 0° — a zero angle would put Q on the line PR, but the stem fixes Q as one of two distinct intersection points, so Q is off that line.
• (d)
  60° — 60° — the angle in a semicircle is fixed at 90° whatever the radius, so no choice of circle sizes can bring ∠ PQR down to 60°.

The angle in a semicircle is the inscribed-angle theorem at its simplest. Arc PR is half the circle, so the angle at the centre is 180°, and any angle standing on that arc from the rest of the circle is half of it — 90°.

The second circle is there to look complicated. Once you see that Q sits on both circles, each diameter drawn from P delivers its own right angle at Q.

A bonus follows. ∠ PQR and ∠ PQS are both 90° and both stand on the ray QP, so R, Q and S lie on one straight line.

Key facts

• An angle subtended by a diameter at any point on the circle is 90°, by Thales' theorem.
• Thales' theorem is the inscribed-angle theorem applied to a central angle of 180°.
• With PR and PS as diameters, ∠ PQR and ∠ PQS are both right angles, placing R, Q and S on one line.

Common traps

• Hunting for the radii, which the question never gives and never needs
• Reading ∠ PQR as an angle of the overlapping lens rather than an angle in circle PR
• Assuming the two circles are equal and halving 90° into 45°

No directly related past PYQ was found.