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: Understanding the Question:
We are given a circle with center $O$.
$PQ$ is a tangent to the circle at point $Q$.
The lengths of the segments are given as $OQ = a$, $OP = a + 2$, and $PQ = 2b$.
We need to find the correct algebraic relation between $a$ and $b$ from the given choices.

Step 2: Key Formula or Approach:
The radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$.
Therefore, $\Delta OQP$ is a right-angled triangle with $\angle OQP = 90^\circ$.
Applying Pythagoras' theorem to $\Delta OQP$:
\[ OP^2 = OQ^2 + PQ^2 \]
We will substitute the given algebraic values into this equation and simplify.

Step 3: Detailed Explanation:

• Set up the Pythagorean equation:
\[ OP^2 = OQ^2 + PQ^2 \]

• Substitute the given values $OQ = a$, $OP = a + 2$, and $PQ = 2b$:
\[ (a + 2)^2 = a^2 + (2b)^2 \]

• Expand both sides of the equation:
\[ a^2 + 4a + 4 = a^2 + 4b^2 \]

• Subtract $a^2$ from both sides to simplify:
\[ 4a + 4 = 4b^2 \]

• Divide the entire equation by 4 to get the simplest relation:
\[ a + 1 = b^2 \implies b^2 = a + 1 \]


Step 4: Final Answer:
The correct algebraic relation between $a$ and $b$ is $b^2 = a + 1$.