AP DSC School Assistant Math Model Test 1

Concept:The radius of a circle is perpendicular to the tangent at the point of contact. In the given figure, OT is the radius and PT is tangent to the circle, so triangle OTP is right‑angled at T. Using the Pythagorean theorem we can relate the known lengths to find $r$.Explanation:Denote the radius $OT = OB = r$ (both are radii of the same quadrant). The length $PB = 36$ cm is the part of the line from the centre O to point P that lies outside the circle. Hence the distance from O to P is $OP = OB + PB = r + 36$ cm.Given that $PT = 96$ cm is the tangent segment, and $\angle OTP = 90^\circ$. Therefore in right‑angled $\triangle OTP$:$$OP^2 = OT^2 + PT^2$$Substitute the values:$$(r + 36)^2 = r^2 + 96^2$$Expand: $r^2 + 72r + 1296 = r^2 + 9216$Cancel $r^2$ from both sides: $72r + 1296 = 9216$So $72r = 9216 - 1296 = 7920$Thus $r = \frac{7920}{72} = 110$ cm.Alternatively, the same result can be obtained using the tangent‑secant theorem: $PT^2 = PB \times PC$, where $PC = PB + 2r = 36 + 2r$. Solving gives $2r = 220$, so $r = 110$ cm.Answer:The radius of the circle is 110 cm, which corresponds to option C.