Circles: MCQs with Answers

A secant intersects a circle at two distinct points.

Hence the required line is a secant.

The radius and tangent form the two legs of a right triangle with hypotenuse OP.

The hypotenuse is longer than either leg.

Hence the tangent length is always less than OP.

A tangent has exactly one common point with a circle.

Hence it touches the circle at one point only.

For intersecting circles, two common tangents can be drawn on the outside.

An internal common tangent would have to pass between the overlapping circles and cannot touch both without crossing a circle.

Hence there are two common tangents.

The perpendicular from the centre to a tangent is the radius through the point of contact.

Radius = 4 cm.

Diameter = 2 × 4 = 8 cm.

Both tangents are perpendicular to the same diameter.

Lines perpendicular to the same line are parallel.

Hence the tangents are parallel.

At a point on a circle, the tangent is perpendicular to the radius through that point.

Only one such perpendicular exists.

Hence exactly one tangent can be drawn.

The tangents lie on opposite sides of the circle and are perpendicular to the diameter joining their points of contact.

Their distance equals the diameter.

Distance = 2 × 5.2 = 10.4 cm.

The perpendicular distance between the parallel tangents equals the diameter.

Distance = 2 × 3 = 6 cm.

A line through a point inside a circle intersects the circle at two points.

It cannot be a tangent.

Thus no tangent can be drawn from an interior point, so the fourth statement is false.

OQ ⟂ PQ, since PQ is a tangent.

PQ = √(OP² − OQ²) = √(17² − 8²) = 15 cm.

Area of triangle OPQ = $\frac{\text{1}}{\text{2}}$ × OQ × PQ.

= $\frac{\text{1}}{\text{2}}$ × 8 × 15 = 60 cm².

OA ⟂ PA and OB ⟂ PB.

PA = √(13² − 5²) = 12 cm.

PA = PB, since tangents from P are equal.

Area of OAPB = 2 × $\frac{\text{1}}{\text{2}}$ × 5 × 12 = 60 cm².

OQ ⟂ PQ and OR ⟂ PR.

PQ = PR = 4 cm, since tangents from P are equal.

Area of PQOR = 2 × $\frac{\text{1}}{\text{2}}$ × 3 × 4.

= 12 cm².

The radius through the point of contact is perpendicular to the tangent.

By Pythagoras, tangent length² = 25² − 7².

Tangent length = √(625 − 49) = 24 cm.

The radius through the point of contact is perpendicular to the tangent.

By Pythagoras, tangent length² = 13² − 5².

Tangent length = √(169 − 25) = 12 cm.

OQ ⟂ PQ.

PQ = √(10² − 6²) = 8 cm.

PR = PQ = 8 cm.

Area of PQOR = 2 × $\frac{\text{1}}{\text{2}}$ × 6 × 8 = 48 cm².

OQ ⟂ PQ.

By Pythagoras, OP² = OQ² + PQ².

(a + 2)² = a² + 4b².

a² + 4a + 4 = a² + 4b².

Hence b² = a + 1.

The tangent is perpendicular to the radius at the point of contact.

In triangle OPT, right-angled at T, OP² = OT² + TP².

41² = 9² + TP².

TP² = 1681 − 81 = 1600.

Hence TP = 40 cm.

The radius is perpendicular to the tangent.

Distance from the centre² = 3² + 4² = 25.

Hence the distance is 5 cm.

PQ = PR = 24 cm.

Area of PQOR = 2 × $\frac{\text{1}}{\text{2}}$ × 24 × 10.

= 240 cm².

Radius = √(13² − 12²).

= √25 = 5 cm.

Distance from centre² = tangent length² + radius².

10² = 8² + r².

r = √(100 − 64).

Hence r = 6 cm.

Join the external point to the centre and one point of contact.

The line to the centre bisects the 60° angle, giving 30°.

The radius is perpendicular to the tangent.

tan 30° = $\frac{\text{5}}{\text{tangent length}}$.

Hence tangent length = 5√3 cm.

The radii are perpendicular to the tangents at the points of contact.

In the resulting quadrilateral, the four angles add to 360°.

Required angle = 360° − 90° − 90° − 130°.

= 50°.

OP ⟂ AP and OQ ⟂ AQ.

In quadrilateral OPAQ, the angle at O is also 90°.

OP = OQ and AP = AQ, so OPAQ is a square.

Hence the radius OP = AP = 2 cm.

Join the external point to the centre and one point of contact.

The line to the centre bisects the 60° angle, giving 30°.

The radius is perpendicular to the tangent.

tan 30° = $\frac{\text{6}}{\text{tangent length}}$.

Hence tangent length = 6√3 cm.

OQ ⟂ PQ.

OP bisects ∠QPR, so ∠QPO = 45°.

sin 45° = $\frac{\text{OQ}}{\text{OP}}$.

$\frac{\text{1}}{\text{√2}}$ = $\frac{\text{4}}{\text{OP}}$.

Hence OP = 4√2 cm.

OQ ⟂ PQ and OR ⟂ PR.

∠QOR = 360° − 90° − 90° − 50° = 130°.

The angle at the centre is twice the angle at T on the remaining part of the circle.

∠QTR = $\frac{\text{130°}}{\text{2}}$ = 65°.

OA ⟂ PA and OB ⟂ PB.

∠AOB = 360° − 90° − 90° − 50° = 130°.

OA = OB, so triangle OAB is isosceles.

∠OAB = $\frac{\text{180° − 130°}}{\text{2}}$ = 25°.

Join the external point to the centre and a point of contact.

The angle at the external point is bisected, giving 30°.

The radius is perpendicular to the tangent.

tan 30° = $\frac{\text{3}}{\text{tangent length}}$.

Hence tangent length = 3√3 cm.

OP ⟂ TP and OQ ⟂ TQ.

The angles of quadrilateral OPTQ add to 360°.

∠PTQ = 360° − 90° − 90° − 110°.

= 70°.

OQ ⟂ PQ and OR ⟂ PR.

∠QOR = 360° − 90° − 90° − 35°.

= 145°.

∠PAO = 90°, because the radius OA is perpendicular to tangent PA.

∠POB is an exterior angle of triangle APO.

By the exterior angle theorem, ∠POB = ∠PAO + ∠APO.

115° = 90° + ∠APO.

Therefore, ∠APO = 25°.

Join OA and OB.

∠OAP = ∠OBP = 90°, since each radius is perpendicular to its tangent.

In quadrilateral AOBP, ∠APB + ∠OAP + ∠AOB + ∠OBP = 360°.

80° + 90° + ∠AOB + 90° = 360°.

Hence ∠AOB = 100°.

The radius to the point of contact is perpendicular to the chord and bisects it.

Half-chord length = √(5² − 4²).

Half-chord length = 3 cm.

Hence chord length = 6 cm.

The perpendicular from the common centre bisects the chord.

Half the chord = 12 cm.

Outer radius² = 12² + 5² = 169.

Hence the outer radius is 13 cm.

OD ⟂ AB, since AB is tangent to the inner circle.

AD = DB = 48 cm, since a perpendicular from the centre bisects a chord.

OD = √(50² − 48²).

= √196 = 14 cm.

OR ⟂ PQ, since PQ is tangent to the inner circle.

PR = RQ = 7 cm.

OR = √(25² − 7²).

= √576 = 24 cm.

The distance of the tangent line from the common centre equals the smaller radius.

This distance is less than the larger radius.

Hence the line is a secant of the larger circle and intersects it at two points.

AB + CD = AD + BC.

AB + 4 = 3 + 7.

AB = 6 cm.

In a tangential quadrilateral, AB + CD = BC + AD.

In a parallelogram, AB = CD and BC = AD.

Thus 2AB = 2BC, so AB = BC.

All four sides are equal.

Hence the parallelogram is a rhombus.

Tangents from each vertex are equal.

Therefore PQ + RS = QR + SP.

12 + 14 = 15 + SP.

Hence SP = 11 cm.