Q) Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

(Q 31A – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s make a diagram for better understanding of the question:

Here, we have a circle with center O and IB is its radius of length R.

We have taken a point A outside the circle at same distance as radius R and from point A, tangent AB is drawn on the circle.

We need to prove that tangent AB is Ʇ to radius OB

or we can say that we need to prove that ∠ OBA is 90 ⁰

Step 2:

Let’s consider that ∠ OBA is 90 ⁰

If our findings do not match with perpendicularity, our assumption will be incorrect, and it willl be proved that ∠ OBA is not 90 ⁰

Step 3: Next, Let’s connect OA and draw a perpendicular on OA from point B.

Now, we compare Δ OBC and Δ ABC:

∴ OB = AB                  (taken initially)

∠ OCB = ∠ ACB         (BC is Ʇ to OA)

BC = BC                    (Common side)

∴ by RHS rule of congruency, Δ OBC Δ ABC

Step 4: Now by CPCT rule,

∠ OBC = ∠ ABC

∵ we considered, ∠ OBA = 90 ⁰

and now ∠ OBC = ∠ ABC

∴ ∠ OBC = ∠ ABC = 45 ⁰

Step 5: Now we look in Δ OBC,

∠ OCB = 90 ⁰

∠ OBC = 45 ⁰

∴ ∠ BOC = 180 ⁰ – 90 ⁰ – 45 ⁰ = 45 ⁰

Step 6: we compare Δ OBC and Δ ABC:

Here, ∠ BOC = ∠ ABC        (= 45 ⁰, from step 4 & 5)

∠ BCA = ∠ BCO                  (BC is Ʇ OA)

∴  Δ OBC ~ Δ ABC

∴

∴  BC ² = OC . AC

Step 7: We look at Right angle triangles, Δ OBC and Δ ABC,

by applying Pythagoras theorem in Δ OBC we get:

OB ² = OC ² + BC ²  …………. (i)

Similarly, from Δ ABC, we get:

AB ² = AC ² + BC ² ………… (ii)

By adding equations (i) and (ii), we get:

OB ² + AB ² = (OC ² + BC ² ) + (AC ² + BC ² )

= OC ² + AC ²  + 2 BC ²

from Step 6, we have BC ² = OC. AC

∴ OB ² + AB ² = OC ² + AC ²  + 2 (OC . AC)

= (OC + AC) ²

∵ OC + AC = OA

∴ OB ² + AB ² = OA ²

This is possible only if ∠ OBA = 90

Since it confirms our assumption, hence ∠ OBA = 90

Therefore, tangent drawn to a point on circle is perpendicular to the radius.

Hence Proved!

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