Q5:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Given: A circle with center $O$. Let $XY$ be a tangent to the circle at the point of contact $P$.

To Prove: The perpendicular to the tangent $XY$ at the point of contact $P$ passes through the center $O$.

Step 1: Assumption for Proof by Contradiction
Let us assume that the perpendicular to the tangent $XY$ at point $P$ does not pass through the center $O$. Instead, let it pass through another point $O'$ such that $O'P \perp XY$.

Step 2: Utilizing the Tangent-Radius Theorem
We know from the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Therefore, the radius $OP$ must be perpendicular to the tangent $XY$.
Mathematically, $OP \perp XY$.
This implies that $\angle OPX = 90^\circ$. [Since $OP$ is the radius at point of contact $P$]

Step 3: Analyzing the Assumption
By our initial assumption in Step 1, we stated that $O'P \perp XY$.
This implies that $\angle O'PX = 90^\circ$.

Step 4: Logical Deduction
Comparing the results from Step 2 and Step 3:
$\angle OPX = 90^\circ$
$\angle O'PX = 90^\circ$
Therefore, $\angle OPX = \angle O'PX$.

Step 5: Conclusion
The equality $\angle OPX = \angle O'PX$ is only possible if the line $O'P$ coincides with the line $OP$.
This means that the point $O'$ must be the same as point $O$.
Thus, our assumption that the perpendicular does not pass through the center is false.

Final Answer: Hence, it is proved that the perpendicular at the point of contact to the tangent to a circle must pass through the center.