Question

$PQ$ is tangent to a circle with centre $O$. If $OQ = a$, $OP = a + 2$ and $PQ = 2b$, then relation between $a$ and $b$ is

• A. $a^2 + (a+2)^2 = (2b)^2$
• B. $b^2 = a + 4$
• C. $2a^2 + 1 = b^2$
• D. $b^2 = a + 1$

Hint:

Always recognize the right angle formed between a tangent and the radius at the point of contact.
This allows immediate application of Pythagoras' theorem!

Answer 1

The Correct Option is D

Step 1: Spot the right angle.
Since $PQ$ is tangent at $Q$, the radius $OQ$ meets it at $90^\circ$, so triangle $OQP$ is right angled at $Q$.
Step 2: Test the answer choices directly using Pythagoras instead of expanding algebra first.
We need $OP^2 = OQ^2 + PQ^2$, i.e. $(a+2)^2 = a^2 + 4b^2$. Try option (4), $b^2 = a+1$, meaning $4b^2 = 4a+4$.
Step 3: Check if it balances.
Right side becomes $a^2 + 4a + 4$, and left side $(a+2)^2 = a^2+4a+4$ too, so they match perfectly.
Step 4: Conclude.
Only option (4) satisfies the Pythagorean condition for every value of $a$.
\[ \boxed{b^2 = a+1} \]