Question:easy

Assertion (A) : Radius is the smallest distance of a tangent from the centre of the circle.
Reason (R) : Radius is perpendicular to the tangent.

Show Hint

To verify if the Reason is the correct explanation, read the two statements by connecting them with "because":
"Radius is the smallest distance of a tangent from the centre of the circle because the radius is perpendicular to the tangent."
Since the shortest distance from a point to a line is always the perpendicular distance, this makes logical and mathematical sense!
Updated On: Jul 7, 2026
  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.
Show Solution

The Correct Option is A

Solution and Explanation

Concept:
This is an Assertion-Reason question on the properties of a tangent to a circle. Instead of just quoting the rule that the perpendicular distance from a point to a line is the shortest distance, let's prove that fact for this exact situation using Pythagoras' theorem, so the reasoning is visible step by step.

Step 1: Check the Reason (R) first.
By the tangent-radius theorem, the radius drawn to the point where a tangent touches a circle is always perpendicular to that tangent. So Reason (R), "Radius is perpendicular to the tangent," is a true statement on its own.

Step 2: Set up a right triangle to test the Assertion (A).
Let $O$ be the centre of the circle, and let the tangent touch the circle at point $P$, so $OP$ is a radius and $OP$ is perpendicular to the tangent at $P$.
Now take any other point $Q$ on the same tangent line, different from $P$.
Since $OP \perp$ tangent, the triangle $OPQ$ has a right angle at $P$.

Step 3: Apply Pythagoras' theorem in triangle OPQ.
In right triangle $OPQ$, right-angled at $P$, the side $OQ$ is the hypotenuse. By Pythagoras' theorem:
\[ OQ^2 = OP^2 + PQ^2 \]
Since $Q$ is a different point from $P$, the length $PQ$ is strictly greater than $0$, so $PQ^2 > 0$. This gives:
\[ OQ^2 = OP^2 + PQ^2 > OP^2 \]
Taking the square root of both sides (both lengths are positive):
\[ OQ > OP \]

Step 4: Interpret this result.
This shows that any point $Q$ on the tangent line, other than the point of contact $P$, is farther from the centre $O$ than $P$ is.
So the distance $OP$, which is exactly the radius, is the smallest possible distance from the centre to any point on the tangent line. This confirms Assertion (A) is true.

Step 5: Link the Reason to the Assertion.
Look back at where the whole argument came from. It only worked because $OP \perp$ tangent let us treat $OPQ$ as a right triangle with $OP$ as one leg and $OQ$ as the hypotenuse. Without that perpendicularity, we could not have used Pythagoras to conclude $OQ > OP$ at all. So the perpendicularity described in Reason (R) is exactly what makes Assertion (A) true, which means R is the correct explanation of A.

Final Answer:
Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A), which corresponds to Option (A).
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