Chapter 9: Circles

False

Justification:

For example,

Consider the given figure. In which we have a circle with centre O and AB a chord with ∠AOB = 60°

Since, tangent to any point on the circle is perpendicular to the radius through point of contact,

We get,

OA ⏊ AC and OB ⏊ CB

∠OBC = ∠OAC = 90° … eq(1)

Using angle sum property of quadrilateral in Quadrilateral AOBC,

We get,

∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°

90° + 90° + 60° + ∠ACB = 360°

∠ACB = 120°

Hence, the angle between two tangents is 120°.

Therefore, we can conclude that,

the given statement is false.

False

Justification:

Length of tangent from an external point P on a circle may or may not be greater than the radius of the circle.

True

Justification:

Consider the figure of a circle with centre O.

Let PT be a tangent drawn from external point P.

Now, Joint OT.

OT ⏊ PT

We know that,

Tangent at any point on the circle is perpendicular to the radius through point of contact

Hence, OPT is a right-angled triangle formed.

We also know that,

In a right angled triangle, hypotenuse is always greater than any of the two sides of the triangle.

Hence,

OP > PT or PT < OP

Hence, length of tangent from an external point P on a circle with centre O is always less than OP.

True

Justification:

The angle between two tangents to a circle may be 0°only when both tangent lines coincide or are parallel to each other.

True.

Tangent is always perpendicular to the radius at the point of contact.

Hence, ∠RPT = 90

If 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.

Consider the following figure,

From point P, two tangents are drawn.

It is given that, OT = a

And line OP bisects ∠RPT.

So,

∠TPO = ∠RPO = 45^o

We know that, OT ⊥ PT

In right-angled triangle OTP,

Sin 45^o = OT/OP

= 1/√2 = a/OP

Hence, OP = a√2