From an external point P, a tangent PT has been drawn to a circle with centre at O and radius 3 cm, intersecting its concentric circle at A and B. If AB = 8 cm and OA = AP, the length PQ equals.
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Concentric circle problems with tangents often contain hidden 3-4-5 right-angled triangles!
Once you identify the perpendicular chord segment as 4 cm and the inner radius as 3 cm, you can immediately identify the outer radius as 5 cm, saving you calculation steps!
Step 1: Spot the right triangle formed by the radius and chord. Since the tangent is perpendicular to the radius at the point of contact, $OQ\perp AB$, and $Q$ bisects chord $AB$, so $AQ=4$ cm with $OQ=3$ cm. Step 2: Recognise the 3-4-5 triple instead of computing a square root. In right triangle $OAQ$ the legs are $3$ and $4$, the classic $3$-$4$-$5$ triple, so the hypotenuse $OA=5$ cm directly. Step 3: Use the given condition $OA=AP$. So $AP=OA=5$ cm. Step 4: Add up along the tangent line. $PQ=PA+AQ=5+4=9$ cm, matching option (C). \[ \boxed{9\text{ cm}} \]
