Circles Class 10 Maths Exercise 10.2 NCERT Solutions

Reading the question correctly. The wording tells you which side is which, so decode the two phrases before drawing anything.

• Distance to centre: the phrase ``distance of Q from the centre'' always means the straight line OQ, and since this line runs from the external point to the centre it is the longest side and therefore the hypotenuse of the right triangle.
• Tangent is a leg: the ``length of the tangent'' is the segment from the external point to the point of contact, which is a leg, and mislabelling these two is the only thing that goes wrong on this question.
• Use the triples: once the hypotenuse is the 25 and the tangent leg is the 24, subtraction gives the radius, and remembering the triples three-four-five, five-twelve-thirteen and seven-twenty-four-twenty-five makes such single-step tangent problems almost instant in the exam.

r=√OQ²-PQ²=√625-576=7 cm, option (A).

Two clean routes to the same answer. You can reach the answer in two ways, and which one to write depends on time and marks.

• Safe route: the quadrilateral method shown above is the most reliable, since it only uses the two right angles at the points of contact and the angle sum of the quadrilateral.
• Fast route: the supplementary rule gives the answer directly as ∠ PTQ=180^∘-110^∘=70^∘, so a student short of time can quote it, while one wanting full working should still set up the quadrilateral on paper.
• The slip to avoid: never treat ∠ POQ as if it equalled ∠ PTQ, because the two are supplementary and not equal, except in the one special case when each of them is a right angle.

∠ PTQ=180^∘-110^∘=70^∘, option (B).

A useful shortcut and a check. The right angle at A lets you skip straight to the answer, then verify it a second way.

• The shortcut: in triangle OAP the angle at A is fixed at a right angle, so the other two angles must add to 90^∘; since ∠ APO is half of 80^∘, the angle ∠ POA is simply 90^∘-40^∘=50^∘ in two lines.
• The check: here ∠ POA equals half of ∠ AOB, which fits the supplementary relation ∠ AOB+∠ APB=180^∘, giving ∠ AOB=100^∘ and ∠ POA=50^∘, exactly the value already found.
• The common slip: students often halve the wrong angle, so always halve the angle between the two tangents and never the angle at the centre.

∠ POA=90^∘-40^∘=50^∘, option (A).

Presenting a proof for full marks. A board proof should open by stating the given and the to-prove in one line, then give a reason beside each step.

• Set it up: write the given as a circle with diameter AB and tangents at A and B, and the to-prove as the claim that these two tangents are parallel to each other.
• Quote the reasons: the two key reasons are Theorem 10.1 that a tangent is perpendicular to the radius, used once at each end, and the geometry fact that two lines perpendicular to the same line are parallel.
• Either ending works: you may instead treat AB as a transversal and point out that the alternate interior angles are both right angles and hence equal, which again forces the tangents to be parallel; either close earns the conclusion mark as long as the perpendicularity is justified first.

Both tangents make 90^∘ with the same diameter, so they are parallel.

Why contradiction is the natural method here. The statement is the converse of Theorem 10.1, and converses are often cleanest when you argue by contradiction.

• The core idea: we already know the radius is one perpendicular to the tangent at P, so if a second perpendicular existed that avoided the centre we would have two perpendiculars at the same point, which breaks a basic uniqueness fact.
• The conclusion: that impossibility forces the perpendicular and the radius to be one and the same line, and since the radius runs to the centre, so does the perpendicular.
• State uniqueness aloud: write the uniqueness of the perpendicular at a point explicitly, because that single sentence is what most answers leave out and lose a mark on.
• Set it up cleanly: open by writing the given, a tangent XY touching at P, and the to-prove, that the perpendicular at P passes through the centre, since a clean setup earns the first method mark; the direct route is harder to phrase at this level, so contradiction stays the recommended method here.

Only one perpendicular exists at P; it is the radius OP, which meets the centre, so the perpendicular passes through O.

Keeping leg and hypotenuse straight. As with the earlier tangent problems, the distance to the centre is always the hypotenuse and the tangent length is always a leg.

• The wrong move: a student who writes r=√5²+4² by mistake gets √41, which is larger than the distance to the centre and therefore clearly impossible to accept as a radius.
• The sanity check: the radius can never exceed the distance from an external point to the centre, so checking that the answer r=3 is less than the distance 5 confirms the work in one glance.
• How to picture it: from the external point, the distance to the centre is the slant or hypotenuse, while the tangent and the radius are the two shorter sides of the triangle.

r=√OA²-AP²=√25-16=3 cm.

Why the chord is bisected. Two separate facts combine here, and the answer is only watertight when you name both of them clearly.

• Chord is a tangent: a chord of the larger circle that touches the inner concentric circle is a tangent to that inner circle, so the radius drawn to the point of contact is perpendicular to the chord at that point.
• Perpendicular bisects: the perpendicular dropped from the centre of a circle to any of its chords always bisects the chord, so the point of contact sits exactly at the mid-point of the chord AB.
• Why it matters: together these two facts put the right angle exactly at the mid-point of AB, which is what lets a single Pythagoras step find half the chord directly from the two radii.
• Finish properly: name both facts in the written answer, since quoting only one leaves the bisection unjustified, and then remember the final doubling that converts the half-length of four into the full chord of eight.

Half-chord =√5²-3²=4 cm, so the full chord AB=8 cm.

A result worth memorising. The conclusion says the sums of opposite sides of a tangential quadrilateral are equal, and it is a standard tool in many later problems.

• Clean layout: label the four points of contact first, then write the four equal tangent pairs, and only then add the equations together to reach the result.
• Cite the theorem: students sometimes lose a mark by not stating which tangent pairs are equal and why, so always quote Theorem 10.2 as the reason each pair is equal.
• Where it returns: this property is exactly why a parallelogram that circumscribes a circle must be a rhombus, the very next result in Q15, because there the equal opposite-side sums force all four sides to be equal in length.

Adding the four equal-tangent pairs from the vertices gives AB+CD=AD+BC.

Reading the angle bisectors correctly. The heart of the proof is that OA and OB are angle bisectors, and everything else follows from that one observation.

• The key step: once a student sees that ∠ OAB+∠ OBA is half of ∠ PAB+∠ QBA, and that this second sum is 180^∘ from the parallel lines, the answer simply drops out of the triangle angle sum.
• Justify the bisector: a common gap is forgetting to say why OA bisects the angle; the reason is the congruence of the two right triangles sharing OA, and stating that congruence keeps the proof watertight and earns the full marks.
• Label first: name the three points of contact, P on the top tangent, Q on the bottom and C on the slant tangent, before writing any equation, because the equal-tangent pairs AP=AC and BQ=BC are what make the bisectors appear.
• Optional confirm: drawing the line OC and noting it is perpendicular to AB confirms the picture, even though the angle answer does not strictly need it, and the final algebra is only a single halving and one subtraction inside the triangle.

OA, OB bisect a co-interior pair summing to 180^∘, so ∠ AOB=180^∘-90^∘=90^∘.

Choosing the quadrilateral carefully. The right figure makes this proof short, and choosing the wrong one makes it stall.

• Pick the quadrilateral: the clean proof uses quadrilateral PAOB with the vertices taken in that cyclic order, so its angle sum is 360^∘ and the two right angles at A and B leave exactly 180^∘ to be shared between the angle at the point and the angle at the centre.
• Why not the triangle: students sometimes try to use triangle OAP on its own, which only relates ∠ APO to ∠ AOP and not the full angles ∠ APB and ∠ AOB, so the single triangle route does not finish the problem and wastes time in the examination.
• Back up the word: the word supplementary must be supported by the equation that sums the two angles to 180^∘, so always write that final line explicitly to claim the concluding mark for the proof.

Quadrilateral PAOB has angle sum 360^∘ with two right angles, leaving ∠ APB+∠ AOB=180^∘.

Building on an earlier result. The smartest answer reuses a result already proved instead of starting from scratch.

• Quote the lemma: cite the tangential-quadrilateral property AB+CD=AD+BC from Q12 rather than re-deriving it with equal tangents, since quoting it saves time and shows that you command the structure of the whole chapter.
• Then pure algebra: from there it is only algebra, as you substitute the two parallelogram equalities into that relation and watch all four sides collapse to a single common length.
• The slow alternative: a student who instead redraws all four points of contact and writes eight tangent equations will still reach the answer but spends far longer, so the lesson is to recognise when a proved lemma can be used directly to turn a long argument into four short lines.

Using AB=CD, AD=BC in AB+CD=AD+BC gives AB=AD, so all sides are equal: a rhombus.

The reliable area method. Two area formulas drive this question, and matching them turns a hard problem into one equation.

• Match two areas: the incircle form Area=r s and Heron's formula in terms of the sides both describe the same triangle, so setting them equal gives one equation in the single unknown x, which is the cleanest route to the sides.
• Cancel safely: the factor (x+14) appears on both sides after squaring, so cancelling it avoids a messy quadratic, but justify that this factor is not zero first, since a side length is always positive.
• Label at the start: naming the equal tangents AF=AE=x is essential, because without one variable for the third vertex the sides cannot be written down and the method stalls before it begins.
• Picture and finish: the contact point D splits the base into eight and six, and the same tangent lengths reappear at the other two vertices, so the whole triangle is built from just three tangent lengths; once x is found every side is a simple sum, and the written answer should still show the squaring and the cancelling so no method mark is lost.

Tangent length x=7, so AB=x+8=15 cm and AC=x+6=13 cm.

Why the angles pair up so neatly. The proof runs on one engine, the equal tangents from each vertex, and the rest is careful counting.

• The engine: the two tangents from any one vertex are equal, which makes the two right triangles at that vertex congruent, and so the two central angles they cut at the centre are equal to each other.
• Halve the turn: with all eight angles around the centre falling into four equal pairs, the full turn of 360^∘ halves to 180^∘ for one angle from each pair, and grouping one side's two angles makes the opposite side take the other two.
• Label the figure: clearly mark the equal pairs on the diagram, because that single labelling carries most of the proof and is exactly what an examiner looks for before awarding the supplementary conclusion to the question.
• The frequent slip: do not assume the four distinct angles are themselves equal, since only the paired angles at each vertex are equal; keeping that distinction clear is the difference between a complete proof and one that quietly assumes the result, so draw the joins to both vertices and contact points first to keep the counting honest.

Equal central-angle pairs from each vertex make opposite sides subtend angles that add to 180^∘.