MP Board · Class 10 · Mathematics · CirclesProve that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. Also, find the length of a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$ if its distance from the centre is evaluated, and calculate the tangents' properties.

Theorem Proof:

Given: A circle with centre $O$, an external point $P$, and two tangents $PA$ and $PB$ touching the circle at points $A$ and $B$ respectively.

To Prove: $\angle APB + \angle AOB = 180^\circ$

Proof:

• Radius-Tangent Perpendicularity: We know that the radius at the point of contact is perpendicular to the tangent. Therefore, $OA \perp PA$ and $OB \perp PB$. This implies that $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
• Quadrilateral Angle Sum Property: Consider the quadrilateral $PAOB$. The sum of all interior angles of a quadrilateral is $360^\circ$. $$\angle APB + \angle PAI (or , \angle OAP) + \angle AOB + \angle OBP = 360^\circ$$
• Substitution: Substituting the values of $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ into the equation: $$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ$$ $$\angle APB + \angle AOB + 180^\circ = 360^\circ$$
• Simplification: $$\angle APB + \angle AOB = 360^\circ - 180^\circ$$ $$\angle APB + \angle AOB = 180^\circ$$
• Hence, the angle between the two tangents and the angle subtended by the line segment joining the points of contact at the centre are supplementary. Proved.

Numerical Problem:

Given data:

• Radius of the circle ($r$) $= 5\text{ cm}$
• Length of the chord ($AB$) $= 8\text{ cm}$

To find:

• The distance of the chord from the centre ($OM$).

Step-by-step Calculation:

1. Let $O$ be the centre of the circle and $AB$ be the chord of length $8\text{ cm}$.
2. Draw a perpendicular $OM$ from the centre $O$ to the chord $AB$.
3. We know that the perpendicular from the centre of a circle to a chord bisects the chord. Therefore: $$AM = MB = \frac{AB}{2} = \frac{8}{2} = 4\text{ cm}$$
4. Join the centre $O$ to the point $A$ to form the radius $OA$. Here, $OA = 5\text{ cm}$.
5. This forms a right-angled triangle $\triangle OMA$, right-angled at $M$ (since $OM \perp AB$).
6. Applying Pythagoras theorem in right $\triangle OMA$: $$OA^2 = OM^2 + AM^2$$
7. Substitute the known values: $$5^2 = OM^2 + 4^2$$ $$25 = OM^2 + 16$$
8. Solve for $OM^2$: $$OM^2 = 25 - 16$$ $$OM^2 = 9$$
9. Take the square root: $$OM = \sqrt{9} = 3\text{ cm}$|

Answer:\nThe distance of the chord from the centre of the circle is $3\text{ cm}$.