Question

AB is a chord of a circle. The length of AB is 24 cm. P is the midpoint of AB. Perpendiculars from P on either side of the chord meet the circle at M and N respectively. If $PM

• A. 17 cm
• B. 18 cm
• C. 19 cm
• D. 20 cm
• E. 21 cm

Hint:

For a chord of length $2a$ and a point $P$ on its perpendicular at distance $d$ from the centre, $r^2=d^2+a^2$. Distances from $P$ to the circle along that line are $r\pm d$.

Answer 1

The Correct Option is B

Step 1: Set up geometry.
Let the centre be $O$, $PO=d$, radius $r$. Since $P$ is midpoint of chord $AB$, $PO\perp AB$ and passes through $O$. Points $M,N$ lie on this line, so \[ PM=r-d,\qquad PN=r+d. \] 

Step 2: Use $PM=8$.
\[ r-d=8 \;\Rightarrow\; r=d+8. \] 

Step 3: Use chord–radius relation.
Half-chord $=12$, hence \[ r^2=d^2+12^2=d^2+144. \] With $r=d+8$: \[ (d+8)^2=d^2+144 \Rightarrow 16d=80 \Rightarrow d=5,\quad r=13. \] 

Step 4: Compute $PN$.
\[ PN=r+d=13+5=18\text{ cm}. \] 

Final Answer: \[ \boxed{\text{(B) 18 cm}} \]

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