NCERT Exemplar Solutions Class 10 Maths Circles — Exercise 9.2

NCERT Exemplar Problems, Class 10 Mathematics — Circles, Exercise 9.2. All 10 true-or-false questions solved.

Ten statements to judge, and justify — the verdict alone earns nothing. Six of the ten are false, which is a higher proportion than usual and a hint about how to read them: most are plausible-sounding generalisations that hold in the picture you happen to draw and fail in general.

Three quantities do most of the work.

The tangent length. From an external point $P$ at distance $d$ from the centre of a circle of radius $r$:

$$PA = \sqrt{d^2 – r^2}$$

It can be made as small as you like by moving $P$ close to the circle, and as large as you like by moving it away.

The angle between the tangents. If the tangents from $P$ meet at angle $2\theta$, then $OP$ bisects that angle and

$$\sin\theta = \frac{r}{OP}, \qquad \text{so} \qquad OP = \frac{r}{\sin\theta}$$

The locus of centres. A circle through two given points has its centre on the perpendicular bisector of the segment joining them. A circle touching a given line at a given point has its centre on the perpendicular to that line at that point. Questions 8 and 9 are exactly this distinction.

A note on the current syllabus. The rationalised Class 10 chapter carries two theorems and nothing else: the tangent is perpendicular to the radius at the point of contact, and the two tangents from an external point are equal. The tangent–chord theorem is absent — “alternate segment” appears nowhere in the chapter — so Questions 7 and 10 are done here by joining the centre instead. The angle in a semicircle, used in Question 10, is Class 9 material and is still taught.

Key insight. When a statement says a quantity is “always” greater or less than another, test the extremes. Move the external point almost onto the circle, or infinitely far away, and see whether the claim survives. Questions 2, 3 and 4 are all settled that way in one line.

Write ‘True’ or ‘False’ and justify your answer in each of the following:

Question 1

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

Solution. Let the tangents at $A$ and $B$ meet at $P$. The four points $P$, $A$, $O$, $B$ form a quadrilateral, and its angles can all be named:

• $\angle OAP = 90^\circ$, since $OA$ is the radius to the point of contact $A$
• $\angle OBP = 90^\circ$, for the same reason at $B$
• $\angle AOB = 60^\circ$, as given

The angles of a quadrilateral sum to $360^\circ$, so

$$\angle APB = 360^\circ – 90^\circ – 90^\circ – 60^\circ = 120^\circ$$

Not $60^\circ$. The statement is false.

The general relation is worth keeping, because it settles a whole family of questions at once:

$$\angle APB = 180^\circ – \angle AOB$$

The angle at the centre and the angle between the tangents are supplementary, not equal. So a chord subtending $60^\circ$ gives tangents at $120^\circ$; a chord subtending $90^\circ$ gives tangents at $90^\circ$ — the only case where the two coincide.

Question 2

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

Solution. The tangent length from an external point at distance $d$ from the centre is

$$PA = \sqrt{d^2 – r^2}$$

and the statement claims this always exceeds $r$. It does not — it can be made as small as you please by taking $P$ close to the circle.

A concrete counterexample. Take $r = 5$ cm and put $P$ just outside the circle at $d = 5.1$ cm. Then

$$PA = \sqrt{5.1^2 – 5^2} = \sqrt{26.01 – 25} = \sqrt{1.01} \approx 1.005 \text{ cm}$$

which is barely a fifth of the radius. Move $P$ closer still and the tangent length shrinks towards zero.

Where the dividing line is. $PA > r$ requires $d^2 – r^2 > r^2$, that is $d > r\sqrt2$. So the statement is true only for external points further than $r\sqrt2$ from the centre, and false for every point between $r$ and $r\sqrt2$. “Always” is what makes it false.

Question 3

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

Solution. Let the tangent touch at $A$. By Theorem 1, $OA \perp PA$, so triangle $OAP$ is right angled at $A$ and $OP$ is its hypotenuse.

The hypotenuse is the longest side of a right-angled triangle, so

$$PA < OP$$

always. The statement is true.

Algebraically it is even clearer:

$$PA = \sqrt{OP^2 – r^2} < \sqrt{OP^2} = OP$$

since $r > 0$ means something strictly positive is being subtracted under the root.

Contrast this with Question 2, which is the reason both appear here. Comparing the tangent with the radius gives a claim that is sometimes true and sometimes false; comparing it with $OP$ gives one that is always true. The difference is that $OP$ is the hypotenuse of the very triangle the tangent belongs to, while the radius is just another leg.

The gap closes but never vanishes: as $P$ moves far away, $PA$ approaches $OP$ but stays strictly below it.

Question 4

The angle between two tangents to a circle may be $0^\circ$.

Solution. True, and there are two ways it happens.

The clean case: parallel tangents. Take the two tangents at the ends of a diameter. Each is perpendicular to that diameter, so the two are perpendicular to the same line and therefore parallel. Parallel lines never meet and the angle between them is $0^\circ$.

Every diameter gives such a pair, so there are infinitely many.

The limiting case. Take two tangents from an external point $P$ and let $P$ move further and further from the circle. The angle $\angle APB$ between them satisfies

$$\sin\left(\frac{\angle APB}{2}\right) = \frac{r}{OP}$$

As $OP$ grows, the right-hand side shrinks towards zero, so the angle does too. The two tangents become more and more nearly parallel.

The statement says “may be”, not “is always”, so a single configuration is enough to make it true — and the diameter construction supplies one exactly, not merely in a limit.

Question 5

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

Solution. Let the tangents touch at $A$ and $B$. The line $OP$ bisects the angle between them, so

$$\angle APO = \frac{90^\circ}{2} = 45^\circ$$

In the right-angled triangle $OAP$ — right angled at $A$ — the radius $OA = a$ is opposite the $45^\circ$ angle and $OP$ is the hypotenuse:

$$\sin 45^\circ = \frac{OA}{OP} \implies \frac{1}{\sqrt2} = \frac{a}{OP} \implies OP = a\sqrt2$$

True.

There is a neater way to see it. With $\angle APB = 90^\circ$ and $\angle OAP = \angle OBP = 90^\circ$, the quadrilateral $OAPB$ has three right angles, so the fourth — $\angle AOB$ — is also $90^\circ$. With $OA = OB = a$ and $PA = PB$, the figure is a square of side $a$, and $OP$ is its diagonal:

$$OP = a\sqrt2$$

The tangent length here is $PA = a$, equal to the radius. That is the boundary case of Question 2: at $d = a\sqrt2$ the tangent exactly equals the radius.

Question 6

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

Solution. Exactly the construction of Question 5 with a different angle. $OP$ bisects, so

$$\angle APO = \frac{60^\circ}{2} = 30^\circ$$

and in the right-angled triangle $OAP$:

$$\sin 30^\circ = \frac{OA}{OP} \implies \frac12 = \frac{a}{OP} \implies OP = 2a$$

The statement claims $a\sqrt3 \approx 1.732a$. The truth is $2a$. False.

Where $a\sqrt3$ comes from, and that is the point of the question. It is the tangent length, not $OP$:

$$PA = \sqrt{OP^2 – a^2} = \sqrt{4a^2 – a^2} = a\sqrt3$$

So the question has swapped one length of the figure for another, and both appear in the same calculation. Reading it carelessly — or remembering “$\sqrt3$ goes with $60^\circ$” as a slogan — produces the wrong verdict.

Compare with Exercise 9.1 Question 9, where a $60^\circ$ pair of tangents to a circle of radius $3$ gave a tangent length of $3\sqrt3$ and a distance $OP = 6$. The same two numbers, and the question there asked for the other one.

Question 7

The tangent to the circumcircle of an isosceles triangle $ABC$ at $A$, in which $AB = AC$, is parallel to $BC$.

Solution. Let $O$ be the circumcentre, so $OA = OB = OC$ are all radii, and let the tangent at $A$ be the line $\ell$.

Step 1 — $OA$ is perpendicular to $BC$. Since $AB = AC$, the point $A$ is equidistant from $B$ and $C$, so it lies on the perpendicular bisector of $BC$. And $O$ is the circumcentre, equidistant from $B$ and $C$, so it lies on that perpendicular bisector too.

Two points determine a line, so the line $OA$ is the perpendicular bisector of $BC$ — and in particular

$$OA \perp BC$$

Step 2 — $OA$ is perpendicular to $\ell$. That is Theorem 1: the tangent at $A$ is perpendicular to the radius $OA$.

Step 3. Two lines perpendicular to the same line are parallel to each other. Since both $BC$ and $\ell$ are perpendicular to $OA$,

$$\ell \parallel BC$$

True.

Most solution sets use the tangent–chord theorem here: the angle between $\ell$ and $AB$ equals $\angle ACB$ in the alternate segment, which equals $\angle ABC$ because the triangle is isosceles, giving equal alternate angles and hence parallel lines. That theorem is not in the current Class 10 textbook; the perpendicular-bisector argument above uses only the two theorems that are.

Question 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 $PQ$.

Solution. The centre of a circle touching the line $PQ$ at $A$ must satisfy one condition: the radius to $A$ is perpendicular to $PQ$, by Theorem 1. So every such centre lies on the perpendicular to $PQ$ at the point $A$.

That is not the perpendicular bisector of $PQ$ — unless $A$ happens to be the mid-point of $PQ$, and nothing in the question says it is. The statement is false.

A picture to fix the distinction. Let $PQ$ run from $(0,0)$ to $(10,0)$ and let $A$ be $(2,0)$. Circles touching $PQ$ at $A$ have centres at $(2, k)$ for any $k \ne 0$ — the vertical line $x = 2$. The perpendicular bisector of $PQ$ is the vertical line $x = 5$. The two lines are parallel and never meet, so no centre of such a circle lies on the perpendicular bisector.

Only when $A$ is the mid-point do the two lines coincide, and then the statement happens to be true. A claim that holds only in a special case is false as stated.

Question 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 $PQ$.

Solution. A circle passing through both $P$ and $Q$ has both of them on its circumference, so both are at a distance of one radius from the centre:

$$OP = OQ = r$$

A point equidistant from $P$ and $Q$ lies on the perpendicular bisector of $PQ$ — that is what the perpendicular bisector is. So every such centre lies on it, and the statement is true.

The converse holds too: every point of the perpendicular bisector is the centre of some circle through $P$ and $Q$, namely the one with radius equal to its distance from $P$. So the perpendicular bisector is exactly the set of such centres, no more and no less.

Read this against Question 8, which is the reason the pair appears together. Both are about loci of centres, and the answers differ because the conditions differ:

• through two points → equidistant from them → perpendicular bisector of the segment
• touching a line at a point → radius perpendicular there → perpendicular to the line at that point

Confusing the two is the mistake both questions are built to catch.

Question 10

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

Solution. Join $O$ to $C$, and work the angles round the figure.

Step 1 — the triangle $ABC$. $AB$ is a diameter and $C$ is on the circle, so $\angle ACB = 90^\circ$ (angle in a semicircle). With $\angle BAC = 30^\circ$, the angle sum gives

$$\angle ABC = 180^\circ – 90^\circ – 30^\circ = 60^\circ$$

Step 2 — the triangle $OBC$ is equilateral. $OB$ and $OC$ are both radii, so the triangle is isosceles with $\angle OCB = \angle OBC = 60^\circ$ — and a triangle with two $60^\circ$ angles has all three, so

$$\angle BOC = 60^\circ$$

Step 3 — the angle $\angle BCD$. $OC$ is the radius to the point of contact and $CD$ is the tangent, so $\angle OCD = 90^\circ$. The angle $\angle OCB$ takes up $60^\circ$ of that, leaving

$$\angle BCD = 90^\circ – 60^\circ = 30^\circ$$

Step 4 — the angle $\angle BDC$. $D$ lies on $AB$ produced beyond $B$, so $\angle CBD$ is the exterior angle of $\angle ABC$ at $B$:

$$\angle CBD = 180^\circ – 60^\circ = 120^\circ$$

and the angle sum in triangle $BCD$ gives

$$\angle BDC = 180^\circ – 120^\circ – 30^\circ = 30^\circ$$

Step 5. In triangle $BCD$ the angles at $C$ and at $D$ are both $30^\circ$, so the triangle is isosceles and the sides opposite them are equal:

$$BC = BD$$

True.

The figure above is drawn to scale, and both segments measure exactly $70$ units against a radius of $70$ — a coincidence of this particular angle, since $\angle BOC = 60^\circ$ makes $BC$ equal to the radius.

Common mistakes

• Question 1 — equating the two angles. The angle at the centre and the angle between the tangents are supplementary, not equal: $\angle APB = 180^\circ – \angle AOB$. They coincide only at $90^\circ$.
• Question 6 — remembering “$\sqrt3$ goes with $60^\circ$” without asking which length. Here $OP = 2a$ and the tangent length is $a\sqrt3$; the question names $OP$. Both numbers come out of the same triangle, so the slogan is no help — write down which side you are computing.
• Questions 2 and 3 — treating “greater than the radius” and “less than $OP$” as the same kind of claim. $OP$ is the hypotenuse of the triangle the tangent belongs to, so the comparison with it is forced; the radius is a different leg, and the comparison with it depends on where $P$ is.
• Question 4 — reading “may be” as “is”. A single configuration makes a “may be” statement true, and the tangents at the ends of a diameter supply one exactly.
• Questions 8 and 9 — confusing the two loci. Circles through two points have centres on the perpendicular bisector of the segment; circles touching a line at a point have centres on the perpendicular at that point. The two coincide only when the point of contact is the mid-point.
• Questions 7 and 10 — reaching for the tangent–chord theorem. It is not in the rationalised Class 10 textbook. Joining the centre gives both results using only the tangent–radius right angle and the isosceles triangle formed by two radii.
• Question 10 — taking $\angle CBD$ as $60^\circ$. $D$ is on $AB$ produced, so $\angle CBD$ is the exterior angle $180^\circ – 60^\circ = 120^\circ$. Using $60^\circ$ makes the angles of triangle $BCD$ fail to sum to $180^\circ$, which is the check that catches it.

Practise next

• Exercises 9.3 and 9.4 — twenty-two questions where these facts are proved rather than judged, including the equal-tangent identities for circumscribed triangles, quadrilaterals and hexagons.
• Exercise 9.1 — worth a second pass, particularly Questions 8 and 9, which are the same tangent-pair configuration as Questions 5 and 6 here with the numbers running the other way.
• Exercise 8.1 — every tangent-and-radius figure is a right-angled triangle, and Questions 5 and 6 here are pure trigonometry once $OP$ bisects the angle.