# NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.1

Exercise 10.1

## Question 1

How many tangents can a circle have?

### Solution

**Video Solution**

**What is Unknown?**

Number of tangents a circle can have

**Reasoning:**

A tangent to a circle is a line that intersects the circle at only one point. On every point on the circle, one tangent can be drawn.

**Steps:**

As per the above reasoning, a circle can have infinitely many tangents.

## Question 2

Fill in the blanks:

(i) A tangent to a circle intersects it in _________ point (s).

(ii) A line intersecting a circle in two points is called a ____________.

(iii) A circle can have _________ parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called _________ .

### Solution

**Video Solution**

**Steps:**

(i) A tangent to a circle intersects it in __ One __ point (s).

**Reasoning:**

A tangent to a circle is a line that intersects the circle at only one point.

(ii) A line intersecting a circle in two points is called a __ Secant __.

**Reasoning:**

Secant is a line that intersects the circle in two points.

(iii) A circle can have ** Two **parallel tangents at the most.

**Reasoning:**

Tangent at any point of a circle is perpendicular to the radius through the point of contact. Extended radius is a diameter which has two end points and hence two tangents which are parallel to themselves and perpendicular to the diameter.

Center \(O,\) diameter \(AB,\text{ tangents}\, PQ, RS\) and, \({PQ}\; \| \;{RS}\)

\(A\) and \(B\) are called as point of contact.

(iv) The common point of a tangent to a circle and the circle is called __ Point of contact __ .

**Reasoning:**

A tangent to a circle is a line that intersects the circle at only one point and that point is called as point of contact.

## Question 3

A tangent \(PQ\) at a point \(P\) of a circle of radius \(\text{5 cm}\) meets a line through the center \(O\) at a point \(Q\) so that \(OQ = \text{12 cm.}\) Length \(PQ\) is:

(A) \(\text{12 cm}\)

(B) \(\text{13 cm}\)

(C) \(\text{8.5 cm}\)

(D) \(\sqrt{119} \,\rm{cm.}\)

### Solution

**Video Solution**

**What is Known?**

Radius \(OP = \text{5 cm} \)

\(OQ = \text{12 cm}\)

**What is Unknown?**

Length of the tangent \(PQ\)

**Reasoning:**

\(\Delta {OPQ}\) is a right-angle triangle according to Theorem** **\(10.1 :\) The tangent at any point of a circle is perpendicular to the radius through the point of contact.

**Steps:**

By Pythagoras theorem

\[\begin{align} {OQ} ^ { 2 } & = {O P} ^ { 2 } + {P Q} ^ { 2 } \\ 12 ^ { 2 } & = 5 ^ { 2 } + {P Q} ^ { 2 } \\ 144 & = 25 + {P Q} ^ { 2 } \\ {P Q} ^ { 2 } & = 119 \\ {P Q} & = \sqrt { 119 } \rm {\;cm } \end{align}\]

The answer is option D.

## Question 4

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

### Solution

**Video Solution**

**What is Known?**

(i) To draw a circle

(ii) Draw one tangent and one secant to the circle parallel to the given line.

**What is Unknown?**

To draw a circle as per known details.

**Steps:**

\(XY\) is the given line.

\(AB\) is the secant parallel to \(XY,\) \({AB}\parallel {XY}\)

\(AQ\) is the tangent parallel to \(XY,\) \({PQ}\parallel {XY}\)

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school