Two circles, centred at P and Q intersect at two points C and D. AB is tangent to the two circles at A and B. If ∠ADB = 68°, then ∠ACB = __________.

Correct — D. Rule: the tangent–chord angle equals the angle in the alternate segment. Apply it once in each circle.

Circle centred at P — AB is the tangent at A, AC the chord, so ∠BAC = ∠ADC.
Circle centred at Q — AB is the tangent at B, BC the chord, so ∠ABC = ∠BDC.

Ray DC lies between DA and DB, so adding the two gives:
∠BAC + ∠ABC = ∠ADC + ∠BDC = ∠ADB = 68°

Triangle ACB closes it, since its three angles total 180°:
∠ACB = 180° − 68° = 112° → option (d)

Why the others are wrong

• (a)
  102° — 102° would leave 78° for ∠BAC + ∠ABC. The alternate-segment pairing pins that sum to ∠ADB, and the question fixes ∠ADB at 68°.
• (b)
  124° — 124° is 180° − 56°. The subtraction is the right move but the number is misread — the given angle is 68°, so the answer is 180° − 68°.
• (c)
  132° — 132° would leave only 48° for the two base angles of triangle ACB. Those are fixed at 68° in total by the tangent–chord rule, so 48° is never available.

Two circles meeting at C and D with a common tangent AB set up two independent alternate-segment relations, one in each circle.

In each circle, the angle between the tangent and a chord from the point of contact equals the inscribed angle that chord subtends in the far segment. Applied at A and at B, the two angles together account for the whole of ∠ADB.

Triangle ACB then finishes the job: its third angle is 180° minus that sum. So in this configuration ∠ACB and ∠ADB are always supplementary, whatever the two radii are.

Key facts

• Tangent–chord (alternate segment) theorem: the angle between a tangent and a chord equals the inscribed angle in the alternate segment.
• For two circles meeting at C and D with a common tangent touching at A and B, ∠ACB + ∠ADB = 180°.
• Here ∠ADB = 68°, so ∠ACB = 112°.
• The supplementary result holds for any radii of the two circles.

Common traps

• Reaching for the tangent–radius right angle, which is true but connects nothing to ∠ADB here.
• Assuming ∠ACB equals ∠ADB because both stand on AB, when they lie in different circles.
• Judging 68° from how the picture is drawn instead of taking it from the statement.

No directly related past PYQ was found.