NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 9: Circles

Exercise 9.2

Step 1: In a circle, when we join the radii \(OA\) and \(OB\) (where \(O\) is the centre), the angle at the centre is given as \(\angle AOB = 60^\circ\).

Step 2: At point \(A\), the tangent is always perpendicular to the radius \(OA\). Similarly, at point \(B\), the tangent is perpendicular to the radius \(OB\).

Step 3: Therefore, the angle formed between the two tangents at points \(A\) and \(B\), say \(\theta\), is related to the angle at the centre by the rule:

\(\theta + \angle AOB = 180^\circ\)

Step 4: Substitute the given value \(\angle AOB = 60^\circ\):

\(\theta + 60^\circ = 180^\circ\)

Step 5: Solve for \(\theta\):

\(\theta = 180^\circ - 60^\circ = 120^\circ\)

Final Answer: The angle between the tangents is \(120^\circ\), not \(60^\circ\). Hence, the given statement is False.

Step 1: Let the centre of the circle be O, radius = r (in SI units: metres, m).

Step 2: Take a point P outside the circle. The distance from the centre O to P is OP (in metres).

Step 3: The length of the tangent PT from point P is given by the formula:

\[ PT = \sqrt{OP^{2} - r^{2}} \]

Step 4: Suppose OP is only a little bigger than r. For example:

• If radius \(r = 10 \, \text{m}\),
• and distance \(OP = 10.1 \, \text{m}\),
• then tangent length \(PT = \sqrt{10.1^{2} - 10^{2}} = \sqrt{102.01 - 100} = \sqrt{2.01} \approx 1.42 \, \text{m}\).

Step 5: Here, the tangent length (≈ 1.42 m) is smaller than the radius (10 m).

Conclusion: The tangent length is not always greater than the radius. It can be smaller. Hence, the statement is False.

Step 1: Let the circle have centre \(O\) and radius \(r\). Take an external point \(P\) outside the circle.

Step 2: Draw a tangent from \(P\) to the circle, touching the circle at point \(T\).

Step 3: In geometry, a radius drawn to the point of tangency is always perpendicular to the tangent. So, \(OT \perp PT\).

Step 4: This makes triangle \(OPT\) a right-angled triangle at \(T\).

Step 5: By Pythagoras theorem: \[ OP^{2} = OT^{2} + PT^{2} \]

Step 6: Substituting, \(OT = r\) (radius), we get: \[ OP^{2} = r^{2} + PT^{2} \]

Step 7: Rearranging: \[ PT^{2} = OP^{2} - r^{2} \]

Step 8: Taking square root: \[ PT = \sqrt{OP^{2} - r^{2}} \]

Step 9: Since \(r^{2} > 0\), we see that: \(PT < OP\).

Final Result: The tangent length from \(P\) (i.e., \(PT\)) is always smaller than the distance \(OP\). Hence the statement is True.

Step 1: A tangent is a line that just touches the circle at one point, without cutting it.

Step 2: If we take any point outside the circle, we can draw exactly two tangents from that point to the circle.

Step 3: These two tangents will meet at the external point and form an angle \(\theta\).

Step 4: This angle is always between \(0^\circ\) and \(180^\circ\).

Step 5: Why not \(0^\circ\)? Because if the angle were \(0^\circ\), both tangents would overlap each other, which means they would become one single line. But then it will no longer be "two tangents".

Step 6: Therefore, the angle can be very small if the point is taken very far away, but it can never be exactly \(0^\circ\).

Final Conclusion: The statement is False.

Step 1: Draw a circle with centre \(O\) and radius \(a\). Take a point \(P\) outside the circle. Draw two tangents from \(P\) that touch the circle at points \(A\) and \(B\).

Step 2: The angle between the tangents at point \(P\) is given as \(90^\circ\). This means \(\angle APB = 90^\circ\).

Step 3: Join \(O\) to \(A\), \(O\) to \(B\), and \(O\) to \(P\). We now have quadrilateral \(OAPB\).

Step 4: The angle at the centre is related to the angle between the tangents. Formula: \(\angle AOB = 180^\circ - \angle APB\). Substituting \(\angle APB = 90^\circ\): \[ \angle AOB = 180^\circ - 90^\circ = 90^\circ. \]

Step 5: Triangle \(AOB\) is isosceles because \(OA = OB = a\) (both are radii). So, \(\angle AOB = 90^\circ\). Each of the other two angles at \(A\) and \(B\) = \(45^\circ\).

Step 6: Consider right triangle \(AOP\). Here, \(OA = a\) (radius), \(\angle AOP = 45^\circ\).

Step 7: Apply the definition of cosine: \[ \cos(\angle AOP) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{OA}{OP} \] Substituting values: \[ \cos 45^\circ = \frac{a}{OP}. \]

Step 8: We know \(\cos 45^\circ = \tfrac{1}{\sqrt{2}}\). So, \[ \frac{1}{\sqrt{2}} = \frac{a}{OP}. \]

Step 9: Cross multiply: \[ OP = a \times \sqrt{2}. \]

Final Result: The distance from the centre to the external point is \[ OP = a\sqrt{2}. \] Hence, the statement is True.

Step 1: Let the angle between the tangents be \(\theta = 60^\circ\).

Step 2: Formula for the distance of the external point \(P\) from the centre \(O\) is:

\( OP = \dfrac{a}{\sin(\theta/2)} \)

where \(a\) is the radius of the circle.

Step 3: Substitute the values:

\( OP = \dfrac{a}{\sin(60^\circ / 2)} = \dfrac{a}{\sin 30^\circ} \)

Step 4: We know \(\sin 30^\circ = 1/2\).

So, \( OP = \dfrac{a}{1/2} = 2a \).

Step 5: The given statement says \( OP = a\sqrt{3} \), but our calculation gives \( OP = 2a \).

Final Answer: The statement is False.

Step 1: Recall the tangent–chord theorem. It says: the angle between a tangent and a chord drawn at the point of contact is equal to the angle made by the chord in the opposite arc of the circle.

Step 2: In \(\triangle ABC\), a tangent is drawn at point \(A\). - The tangent at \(A\) and the chord \(AB\) make an angle. By the theorem, this angle is equal to \(\angle ACB\). - Similarly, the tangent at \(A\) and the chord \(AC\) make another angle. By the theorem, this angle is equal to \(\angle ABC\).

Step 3: Since the triangle is isosceles (\(AB = AC\)), the base angles are equal. So, \(\angle ABC = \angle ACB\).

Step 4: That means the tangent at \(A\) makes equal angles with \(AB\) and \(AC\). But these are exactly the same angles that the base \(BC\) makes with \(AB\) and \(AC\).

Step 5: Therefore, the tangent line at \(A\) is parallel to the base \(BC\).

Final Answer: The statement is True.

Step 1: Recall the property of a tangent. A circle touches a line at only one point. The radius drawn to the point of contact is always perpendicular to the tangent at that point.

Step 2: Here, the line segment is \(PQ\). The circle touches it at a point \(A\). So the centre of the circle must lie on the line passing through \(A\) and perpendicular to \(PQ\).

Step 3: The perpendicular bisector of \(PQ\) is a special line. It is perpendicular to \(PQ\) and passes through the midpoint of \(PQ\).

Step 4: For the circle’s centre to lie on the perpendicular bisector of \(PQ\), the point of contact \(A\) must be exactly the midpoint of \(PQ\). But in general, \(A\) can be anywhere on \(PQ\), not necessarily at the midpoint.

Step 5: Therefore, the centre of such circles lies on the line through \(A\) perpendicular to \(PQ\), not on the perpendicular bisector of \(PQ\) (except in the special case when \(A\) is the midpoint).

Step 1: Think of a circle. The centre of a circle is always the same distance from every point on the circle.

Step 2: Here, every circle passes through points \(P\) and \(Q\). So, the centre of the circle must be equal distance (equidistant) from \(P\) and from \(Q\).

Step 3: The set of all points that are equal distance from \(P\) and \(Q\) forms a special line. That line is called the perpendicular bisector of \(PQ\).

Step 4: Therefore, the centre of each circle through \(P\) and \(Q\) must lie on this perpendicular bisector.

Step 5: So, the statement is True.

Step 1: We are given that \(AB\) is a diameter of the circle.

From the circle theorem, the angle made by a diameter at the circumference is always a right angle (\(90^\circ\)).

Therefore, in triangle \(ABC\), \(\angle ACB = 90^\circ\).

Step 2: It is also given that \(\angle BAC = 30^\circ\).

Since the sum of angles in a triangle is \(180^\circ\):

\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\)

\(30^\circ + \angle ABC + 90^\circ = 180^\circ\)

\(\angle ABC = 60^\circ\).

Step 3: Now look at the tangent at point \(C\). By the Tangent–Chord Theorem (also called Alternate Segment Theorem):

The angle between the tangent at \(C\) and the chord \(BC\) is equal to the angle in the opposite arc, which is \(\angle BAC\).

So, \(\angle BCD = \angle BAC = 30^\circ\).

Step 4: Now consider triangle \(BCD\):

- \(\angle BCD = 30^\circ\) (from Step 3)
- \(\angle CBD = 60^\circ\) (this is the same as \(\angle ABC\) from Step 2, since point D lies on the extension of AB).

Thus in triangle \(BCD\), the two angles are:

\(\angle CBD = 60^\circ\), \(\angle BCD = 30^\circ\).

Step 5: This makes triangle \(BCD\) a right triangle with angles 90°, 60°, and 30°.

In a 30°–60°–90° triangle, the sides opposite 30° and 60° have a fixed ratio. But here, notice something special: the sides opposite equal angles are equal.

In \(\triangle CBD\), angles at \(C\) and \(D\) are equal (each is 30°). So, the sides opposite them are equal.

That means \(BC = BD\).

Final Conclusion: The statement is True. We proved that \(BC = BD\).