MP Board · Class 10 · Mathematics · CirclesProve that the lengths of tangents drawn from an external point to a circle are equal. Also, apply this theorem to find the length of the tangent segment from a point $Q$ at a distance of $25\text{ cm}$ from the centre of a circle of radius $7\text{ cm}$.

Theorem Proof:

Given: A circle with centre $O$, an external point $P$, and two tangents $PQ$ and $PR$ drawn from $P$ to the circle touching at points $Q$ and $R$ respectively.

To Prove: $PQ = PR$

Construction: Join $OP$, $OQ$, and $OR$.

Proof:

• We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$.
• Consider the two right-angled triangles $\triangle OQP$ and $\triangle ORP$.
• In these triangles:
  • $OQ = OR$ (Radii of the same circle)
  • $OP = OP$ (Common hypotenuse)
  • $\angle OQP = \angle ORP = 90^\circ$ (Right angles)
• Therefore, by RHS (Right angle-Hypotenuse-Side) congruence criterion, $\triangle OQP \cong \triangle ORP$.
• Consequently, by Corresponding Parts of Congruent Triangles (CPCT), we get $PQ = PR$.
• Hence, proved.

Numerical Problem:

Given data:

• Radius of the circle ($r$ or $OQ$) $= 7\text{ cm}$
• Distance of the external point from the centre ($OP$) $= 25\text{ cm}$

To find:

• Length of the tangent ($PQ$)

Step-by-step Calculation:

1. By theorem, the radius is perpendicular to the tangent at the point of contact. Therefore, $\triangle OQP$ is a right-angled triangle, right-angled at $Q$ ($\angle OQP = 90^\circ$).
2. Applying Pythagoras theorem in right $\triangle OQP$: $$OP^2 = OQ^2 + PQ^2$$
3. Substitute the given values into the equation: $$25^2 = 7^2 + PQ^2$$ $$625 = 49 + PQ^2$$
4. Rearrange to solve for $PQ^2$: $$PQ^2 = 625 - 49$$ $$PQ^2 = 576$$
5. Take the square root on both sides: $$PQ = \sqrt{576} = 24\text{ cm}$|

Answer:\nThe length of the tangent is $24\text{ cm}$.