Question

The line joining the centre of a circle to the mid-point of a chord is always:

• A. parallel to the chord
• B. perpendicular to chord
• C. equal to the chord
• D. tangent to the chord

Answer 1

The correct option is B

perpendicular to chord

Explanation for the correct option:

Step 1: Given data:

First, we will draw the diagram,

Here, in the given diagram, $OA$and $OB$ are the radius of the triangle.

Point $O$ is the centre and $AB$ is the chord of the circle.

Let $D$ be the mid-point of $AB.$

Step 2: Congruency of the triangle

Now in triangle $△AOD$ and $△BOD$.

We have,

$OB=OA$ (As the radius of the circle are equal)
As $D$ be the mid-point of $AB.$

So, $AD=BD$

$OD=OD$ (Common)
Hence, $△AOD\cong △BOD$ by $SSS$ criteria.
So, the corresponding angle will be also equal.

So, $\angle ODA=\angle ODB$
Step 3: Angle sum property of triangle

As we know,

The sum of all the angles on one side of a straight line is always $180°$

So,

$\begin{array}{rcl}\angle ODA+\angle ODB& =& 180°\\ \angle ODA+\angle ODA& =& 180°\\ 2\angle ODA& =& 180°\\ \angle ODA& =& \frac{180°}{2}=90°\end{array}$

Hence, $OD\perp AB$

Hence, The line joining the centre of a circle to the mid-point of a chord is always perpendicular to the chord.

Hence, Option (B) is the correct option.