JEE Mathematics Circles MCQs Set 18

Answer: (c) inside the triangle

Explanation:
1. The circumcentre is the point where the perpendicular bisectors of the sides meet.
2. In an acute-angled triangle all three angles are less than \( 90^\circ \), so the circumcentre falls inside the triangle.
3. (It lies on the hypotenuse for a right-angled triangle and outside for an obtuse-angled triangle.)

Teacher's Note:
1. The position of the circumcentre depends on the type of triangle: inside for acute, on the hypotenuse for right, outside for obtuse.
2. Remember this three-way rule for quick reasoning questions.

Answer: (a) both diameters

Explanation:
1. If chords AB and CD bisect each other, then ACBD is a parallelogram.
2. A parallelogram inscribed in a circle is a rectangle, so its diagonals AB and CD are equal and each passes through the centre.
3. Hence both chords are diameters.

Teacher's Note:
1. A cyclic parallelogram is always a rectangle, because opposite angles of a cyclic quadrilateral are supplementary.
2. Chords that bisect each other can only meet at the centre.

Answer: (d) acute (less than \( 90^\circ \))

Explanation:
1. The angle in a major segment is half of the angle subtended by the minor arc at the centre.
2. The minor arc subtends an angle less than \( 180^\circ \) at the centre, so the angle in the major segment is less than \( 90^\circ \).

Teacher's Note:
1. Angle in a semicircle is exactly \( 90^\circ \); in a major segment it is smaller, and in a minor segment it is larger.
2. Use the arc-centre relation: angle at centre = 2 x angle at circumference.

Answer: (b) every diameter

Explanation:
1. Reflecting the circle in any line through its centre maps every point of the circle to a point at the same distance from the centre.
2. So the circle is symmetrical about every diameter.

Teacher's Note:
1. A circle has infinitely many lines of symmetry, one for each diameter.
2. This is why equal chords are equidistant from the centre.

Answer: (b) bisects \( \angle CAB \)

Explanation:
1. In triangles AOB and AOC: AB = AC, OB = OC (radii) and AO is common.
2. So the triangles are congruent (SSS), giving \( \angle OAB = \angle OAC \).
3. Hence AO bisects \( \angle CAB \).

Teacher's Note:
1. Equal chords are equidistant from the centre, so the centre lies on the bisector of the angle between them.
2. Note that AO is also perpendicular to BC.

Answer: (d) \( \frac{27\pi}{4} \) sq. cm

Explanation:
1. Radius of the ball = 3 cm. The water surface is at a distance \( 3 - 1.5 = 1.5 \) cm from the centre.
2. Radius of the cross-section: \( r^2 = 3^2 - 1.5^2 = 9 - 2.25 = 6.75 \).
3. Area \( = \pi r^2 = 6.75\pi = \frac{27\pi}{4} \) sq. cm.

Teacher's Note:
1. The surface circle is a chord-circle of the sphere, so use Pythagoras with the distance from the centre.
2. Be careful to subtract the immersed depth from the radius to get the centre distance.

Answer: (a) \( TA \times TB \)

Explanation:
1. Triangles TPA and TBP are similar (common angle at T and the tangent-chord angle equals the angle in the alternate segment).
2. So \( \frac{TP}{TB} = \frac{TA}{TP} \), giving \( TP^2 = TA \cdot TB \).

Teacher's Note:
1. This is the tangent-secant theorem (power of a point).
2. It also gives equal tangent lengths when two tangents are drawn from T.

Answer: (c) supplementary

Explanation:
1. Opposite angles of a cyclic quadrilateral stand on arcs that together make the whole circle.
2. So their sum is \( 180^\circ \); they are supplementary.

Teacher's Note:
1. Only in a cyclic rectangle or square are opposite angles also equal.
2. The converse also holds: a quadrilateral with supplementary opposite angles is cyclic.

Answer: (c) \( \frac{AB}{\sqrt{2}} \)

Explanation:
1. Let O be the centre and r the radius, so AB = 2r and \( \angle AOC = 90^\circ \).
2. In right triangle AOC: \( AC^2 = r^2 + r^2 = 2r^2 \), so \( AC = r\sqrt{2} \).
3. Since \( r = \frac{AB}{2} \), \( AC = \frac{AB}{\sqrt{2}} \).

Teacher's Note:
1. AC is the side of the square ABCD inscribed in the circle.
2. The side of an inscribed square is radius times root 2.

Answer: (a) 7 cm

Explanation:
1. The centre and any two adjacent vertices form an equilateral triangle, since each central angle is \( 60^\circ \) and the two radii are equal.
2. So the radius equals the side, 7 cm.

Teacher's Note:
1. In a regular hexagon, circumradius = side.
2. The inradius is \( \frac{\sqrt{3}}{2} \) times the side.

Answer: (d) outside the triangle

Explanation:
1. For an obtuse angle at A, the altitudes from B and C meet the opposite sides only when those sides are extended.
2. Hence the altitudes meet outside the triangle.

Teacher's Note:
1. Acute triangle: orthocentre inside; right triangle: at the right-angle vertex; obtuse triangle: outside.
2. Do not confuse it with the centroid, which is always inside.

Answer: (b) incentre

Explanation:
1. Points on the bisector of an angle are equidistant from the two arms of the angle.
2. The point where the three angle bisectors meet is therefore equidistant from all three sides; it is the incentre.

Teacher's Note:
1. The circumcentre is equidistant from the vertices, not the sides.
2. The incentre is the centre of the inscribed circle.

Answer: (b) coincide at a single point

Explanation:
1. In an equilateral triangle each median is also an angle bisector, an altitude and a perpendicular bisector.
2. So the points where these lines meet are the same point.

Teacher's Note:
1. This common point is also the orthocentre.
2. Only in an equilateral triangle do all four centres coincide.

Answer: (d) the interior opposite angle

Explanation:
1. The exterior angle and its adjacent interior angle form a linear pair, so they add up to \( 180^\circ \).
2. The opposite interior angle is also supplementary to that adjacent angle (cyclic quadrilateral).
3. Therefore the exterior angle equals the interior opposite angle.

Teacher's Note:
1. This property is often used to prove that four points are concyclic.
2. Do not confuse it with the exterior angle of a triangle.

Answer: (a) \( \sqrt{d_1 d_2} \)

Explanation:
1. Radii are \( r_1 = \frac{d_1}{2} \) and \( r_2 = \frac{d_2}{2} \); the distance between centres is \( r_1 + r_2 \).
2. Length of a direct common tangent \( = \sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} = 2\sqrt{r_1 r_2} \).
3. \( 2\sqrt{\frac{d_1}{2}\cdot\frac{d_2}{2}} = \sqrt{d_1 d_2} \).

Teacher's Note:
1. For circles touching externally the direct common tangent is twice the geometric mean of the radii.
2. Drop a perpendicular from the smaller centre to the larger radius to see the right triangle.

Answer: (c) 18 cm

Explanation:
1. By the tangent-secant theorem \( AT^2 = AB \cdot AC \).
2. \( 144 = 6 \cdot AC \), so \( AC = 24 \) cm.
3. \( BC = AC - AB = 24 - 6 = 18 \) cm.

Teacher's Note:
1. The product of the whole secant and its external part equals the square of the tangent.
2. A common slip is to stop at AC = 24 cm; the question asks for BC.

Answer: (c) touch each other internally

Explanation:
1. Difference of the radii \( = 10 - 8 = 2 \) cm, which equals the distance between the centres.
2. Distance between centres equal to the difference of radii means internal contact.

Teacher's Note:
1. Compare the centre distance d with r1 + r2 and |r1 - r2|.
2. d = r1 + r2: external contact; d = |r1 - r2|: internal contact; in between: two intersection points.

Answer: (a) parallel to each other

Explanation:
1. A tangent is perpendicular to the radius at the point of contact.
2. Both tangents are perpendicular to the same line (the diameter), so they are parallel.

Teacher's Note:
1. Parallel lines never intersect, so no tangent pair at the ends of a diameter meets.
2. The converse is also true for parallel tangents.

Answer: (d) outside the segment

Explanation:
1. A minor segment is cut off by a chord and lies on the side of the chord away from the centre.
2. Hence the centre is outside the minor segment.

Teacher's Note:
1. The centre lies on the chord only when the chord is a diameter, giving two semicircular segments.
2. The centre is inside the major segment.

Answer: (b) \( 40^\circ \)

Explanation:
1. Opposite angles of a cyclic quadrilateral are supplementary: \( \angle ABC = 180^\circ - 130^\circ = 50^\circ \).
2. \( \angle ACB = 90^\circ \) (angle in a semicircle).
3. In triangle ABC: \( \angle BAC = 180^\circ - 90^\circ - 50^\circ = 40^\circ \).

Teacher's Note:
1. Look for the diameter first; it hands you a right angle.
2. Then use the cyclic quadrilateral property to find the other angle.