Circle: Tangents and Intersecting Chords - Class 10 MCQ

The line OP is drawn from the centre O to point P on the circle. So OP is the radius.

A tangent at a point on the circle is always perpendicular to the radius drawn to that point.

Therefore, ∠OPA = 90°.

This is a standard circle property: Radius ⟂ Tangent.

Hence, Option C is correct.

From an external point, two tangents drawn to a circle are always equal in length.

That means PA = PB.

Given PA = 12 cm.

Therefore, PB must also be 12 cm.

Hence, Option C is correct.

When two circles touch externally, the distance between their centres equals the sum of their radii.

Here radius of first circle r₁ = 5 cm, and of second circle r₂ = 3 cm.

Distance = r₁ + r₂ = 5 + 3 = 8 cm.

Hence, the distance between the centres = 8 cm.

So, Option C is correct.

When two circles touch internally, the distance between their centres equals the difference of their radii.

Here radius of larger circle r₁ = 9 cm, radius of smaller circle r₂ = 6 cm.

Distance = |r₁ − r₂| = |9 − 6| = 3 cm.

Therefore, the centres are 3 cm apart.

Hence, Option A is correct.

Theorem: If two chords intersect inside a circle, then the product of the segments of one chord = product of the segments of the other chord.

i.e., PA × PB = PC × PD.

Substituting the values: (4 × 6) = (3 × PD).

Left side = 24. So, 24 = 3 × PD.

Divide both sides by 3: PD = 24 ÷ 3 = 8.

Therefore PD = 8 cm.

Hence, Option D is correct.

Use the tangent-secant theorem: (tangent length)² = (external part of secant) × (whole secant).

External part = PA = 4 cm; whole secant = PB = 12 cm.

So PT² = PA × PB = 4 × 12 = 48.

PT = √48 = √(16 × 3) = 4√3 ≈ 6.928, but the nearest clean choice is 6 (Option C) — however exact PT = 4√3 cm.

Under typical multiple-choice rounding to integer, Option C (6 cm) is intended as closest simple option.

In right triangle OTP, OP is hypotenuse, OT is radius perpendicular to tangent PT (OT ⟂ PT).

By Pythagoras: PT² = OP² − OT².

OP = 26 cm, OT = 10 cm, so PT² = 26² − 10² = 676 − 100 = 576.

PT = √576 = 24 cm.

Hence Option A is correct.

For an external point P, angle between tangents ∠APB = 180° − angle subtended at centre by chord AB (i.e., ∠AOB).

(Reason: Quadrilateral AOPB has ∠AOP + ∠APB + ∠PBO + ∠OAB relations; known result gives ∠APB = 180° − ∠AOB.)

Rearranged: ∠AOB = 180° − ∠APB.

Substitute ∠APB = 80° → ∠AOB = 180° − 80° = 100°.

Wait — check standard relation carefully: Actually angle between tangents = 180° − central angle. So ∠AOB = 180° − 80° = 100°.

Therefore correct central angle is 100°; Option B (100°) is the correct precise choice.

For two secants from external point P: (external part1) × (whole secant1) = (external part2) × (whole secant2).

For chord through A and B: external part = PA = 2, whole secant = PB = 8 (given). Product = 2 × 8 = 16.

For the other secant through C and D: external part = PC = 4, whole secant = PD (unknown). So 4 × PD = 16.

Solve PD = 16 ÷ 4 = 4.

So PD = 4 cm — Option B is the exact answer.

The perpendicular from centre O to chord AB meets AB at M (midpoint), so AM = BM = 10/2 = 5 cm.

OM = 6 cm, OA = radius r. Right triangle OMA: OA² = OM² + AM².

So r² = 6² + 5² = 36 + 25 = 61.

r = √61 ≈ 7.81 cm; exact value is √61. None of the options show √61, but Option A (8 cm) is the closest simple value.

Correct exact radius = √61 cm; nearest integer option A (8 cm) chosen.

Theorem (angle between tangent and chord): Angle between tangent at A and chord AB equals angle in the opposite arc subtended by chord AB.

Given angle between tangent and chord at A = 50°.

Therefore angle subtended by chord AB at any point C on the opposite arc (∠ACB) = 50°.

Hence Option A is correct.

When two circles touch externally at a point P, their radii to P are collinear because both radii are perpendicular to the common tangent at P.

Thus O1P and O2P lie on the same straight line through P, i.e., O1, P, O2 are collinear.

Therefore Option A is correct.

In triangle OPT, OT = radius = 15 cm, PT = tangent length = 20 cm, OP is hypotenuse.

By Pythagoras: OP² = OT² + PT² = 15² + 20² = 225 + 400 = 625.

OP = √625 = 25 cm.

So Option A (25 cm) is correct.

Product of segments theorem (internal intersection): AE × EB = CE × ED.

AE × EB = 3 × 9 = 27. So CE × ED = 27.

CE = 6, so 6 × ED = 27 → ED = 27 ÷ 6 = 4.5 cm.

That is 4.5 cm — not listed. The closest option in integers is none; Option A (4 cm) is closest but not exact.

Exact ED = 4.5 cm. If forced to pick from given integers, Option A would be nearest but not exact.

Half chord = 12/2 = 6 cm. OM (perpendicular from O to chord) = 5 cm. OA = r. Right triangle OMA gives r² = 5² + 6² = 25 + 36 = 61.

r = √61 ≈ 7.81 cm. None of options show √61. However Option A (13 cm) equals √169, not √61.

Exact radius = √61 cm (≈7.81). Since options do not include correct value, Option A (13 cm) is incorrect. There is no correct option listed.

Exact answer: r = √61 cm.

Step 1: Recall the property: The length of the tangent from an external point P to the circle is given by √(OP² - r²), where OP is the distance from center to P, and r is the radius.
Step 2: Here, OP = 13 cm, r = 5 cm.
Calculation: Length = √(13² - 5²) = √(169 - 25) = √144 = 12 cm.
Conclusion: The tangent length is 12 cm
So Option C correct.

Step 1: From Corollary: The two tangents from an external point are equal in length, say PA = PB = l.
Step 2: The product is PA × PB = l × l = l².
Calculation: l = 15 cm, so product = 15² = 225 cm².
Conclusion: The product is 225 cm²
Option A correct.

Step 1: Theorem 11 (External Touching): Distance d between centers = r1 + r2.
Step 2: r1 = 4 cm, r2 = 6 cm.
Calculation: d = 4 + 6 = 10 cm.
Conclusion: Distance is 10 cm,
Option C correct.

Step 1: Theorem 11 (Internal Touching): Distance d = |r1 - r2|.
Step 2: r1 = 8 cm, r2 = 3 cm (assuming r1 > r2).
Calculation: d = 8 - 3 = 5 cm.
Conclusion: Distance is 5 cm, making Option A correct.

Step 1: Theorem 12 (External Intersection): PA × PB = PC × PD.
Step 2: Here, for external, it's the whole segments: PA × PB = PC × PD, but PA and PB are segments from P to A and A to B?
Standard: if AB is chord, P outside, then PA × PB = PC × PD, where PB = PA + AB, etc.
Calculation: 10 × 3 = 5 × PD, 30 = 5 PD, PD = 6 cm.
Conclusion: PD = 6 cm