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NCERT Exemplar · Class 10 Mathematics Circles

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EXERCISE 9.2 1–10 (part 2 of 5)

  1. Write 'True' or 'False' and justify your answer in each of the following :

    Exercise 1

    If a chord AB subtends an angle of 60\displaystyle 60° at the centre of a circle, then angle between the tangents at A and B is also 60∘\displaystyle 60^{\circ}.

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    NCERT’s answer
    False
    False. NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q1 \[\angle OAP = \angle OBP = 90^{\circ} \quad \text{(radius \(\displaystyle \perp\) tangent)} \] \[\angle AOB + \angle APB = 360^{\circ} - 90^{\circ} - 90^{\circ} = 180^{\circ} \] \[\angle APB = 180^{\circ} - 60^{\circ} = 120^{\circ} \neq 60^{\circ} \]

  2. Exercise 2

    The length of tangent from an external point on a circle is always greater than the radius of the circle.

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    NCERT’s answer
    False
    False. NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q2 \[PA = \sqrt{OP^{2} - r^{2}} \quad \text{(right triangle \(\displaystyle OAP\))} \] \[PA < r \iff OP < r\sqrt{2} \] So for any external point with \(\displaystyle r < OP < r\sqrt{2}\), the tangent is shorter than the radius.

  3. Exercise 3

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

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    NCERT’s answer
    True
    True. NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q3 \[\angle OAP = 90^{\circ} \quad \text{(radius \(\displaystyle \perp\) tangent)} \] \[OP^{2} = OA^{2} + PA^{2} \quad \text{(Pythagoras)} \] \[OP > PA \quad \text{(hypotenuse is the longest side)} \]

  4. Exercise 4

    The angle between two tangents to a circle may be 0\displaystyle 0°.

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    NCERT’s answer
    True
    True. NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q4 \[l \perp AB, \quad m \perp AB \quad \text{(radius \(\displaystyle \perp\) tangent, \(\displaystyle A,O,B\) collinear)} \] \[l \parallel m \] Tangents at the two ends of a diameter are parallel: the angle between them is \(\displaystyle 0^{\circ}\).

  5. Exercise 5

    If angle between two tangents drawn from a point P to a circle of radius a\displaystyle a and centre O is 90∘\displaystyle 90^{\circ}, then OP=a2\displaystyle \mathrm{OP}=a \sqrt{2}.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q5 \[\angle OAP = \angle OBP = 90^{\circ} \quad \text{(radius \(\displaystyle \perp\) tangent)} \] \[\angle APB = 90^{\circ} \quad \text{(given)} \] \[\angle APO = \tfrac{1}{2}\angle APB = 45^{\circ} \quad \text{(\(\displaystyle OP\) bisects \(\displaystyle \angle APB\))} \] \[\sin 45^{\circ} = \dfrac{OA}{OP} = \dfrac{a}{OP} \] \[OP = \dfrac{a}{\sin 45^{\circ}} = a\sqrt{2} \]

  6. Exercise 6

    If angle between two tangents drawn from a point P to a circle of radius a\displaystyle a and centre O is 60∘\displaystyle 60^{\circ}, then OP=a3\displaystyle \mathrm{OP}=a \sqrt{3}.

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    NCERT’s answer
    False
    False. \[\angle OPA = \tfrac{1}{2}(60^\circ) = 30^\circ \quad \text{(OP bisects the angle between the tangents)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q6 \[\sin(\angle OPA) = \frac{OA}{OP} = \frac{a}{OP} \] \[\sin 30^\circ = \frac{a}{OP} \implies OP = 2a \] \[2a \neq a\sqrt{3} \]

  7. Exercise 7

    The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}, is parallel to BC.

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    NCERT’s answer
    True
    True. \[\angle TAB = \angle ACB \quad \text{(tangent–chord angle = angle in alternate segment)} \] \[\angle ACB = \angle ABC \quad \text{(} AB = AC \text{, base angles of isosceles } \triangle ABC \text{)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q7 \[\angle TAB = \angle ABC \implies TA \parallel BC \quad \text{(alternate angles, transversal } AB \text{)} \]

  8. Exercise 8

    If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

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    NCERT’s answer
    False
    False. \[OA \perp PQ, \quad O'A \perp PQ \quad \text{(radius} \perp \text{tangent)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q8 So every centre lies on the line through \(\displaystyle A\) perpendicular to \(\displaystyle PQ\). That line is the perpendicular bisector only if \(\displaystyle A\) is the midpoint \(\displaystyle M\) of \(\displaystyle PQ\); \(\displaystyle A\) need not be (in the figure, \(\displaystyle PA \neq AQ\)).

  9. Exercise 9

    If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. \[CP = CQ \quad \text{(radii of the same circle, for every such centre } C \text{)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q9 \[CP = CQ \implies C \in \text{perpendicular bisector of } PQ \quad \text{(locus of points equidistant from } P, Q \text{)} \]

  10. Exercise 10

    AB is a diameter of a circle and AC is its chord such that ∠BAC=30∘\displaystyle \angle \mathrm{BAC}=30^{\circ}. If the tangent at C intersects AB extended at D, then BC=BD\displaystyle \mathrm{BC}=\mathrm{BD}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. \[\angle ACB = 90^\circ \quad \text{(angle in a semicircle, } AB \text{ diameter)} \] \[\angle ABC = 180^\circ - 90^\circ - 30^\circ = 60^\circ \] \[OB = OC \quad \text{(radii)} \implies \angle OCB = \angle OBC = \angle ABC = 60^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-2_Q10 \[\angle OCD = 90^\circ \quad \text{(tangent} \perp \text{radius at } C \text{)} \] \[\angle BCD = \angle OCD - \angle OCB = 90^\circ - 60^\circ = 30^\circ \] \[\angle DBC = 180^\circ - \angle ABC = 120^\circ \quad \text{(} A, B, D \text{ collinear)} \] \[\angle BDC = 180^\circ - 120^\circ - 30^\circ = 30^\circ \] \[\angle BDC = \angle BCD \implies BC = BD \quad \text{(isosceles } \triangle BCD \text{)} \]