Circles — Practice Quiz
Karnataka Board · Class 10 · Mathematics
Try a 4-question quiz on Circles for Karnataka Board Class 10 Mathematics: tap an answer to check it and see why. 45 questions in the full chapter test.
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Quick Quiz: Circles
0/4Tap an answer to check it instantly. No sign-up needed for these 4.
A tangent to a circle intersects the circle at how many points?
From a point P outside a circle, PA and PB are two tangents drawn to the circle. If PA = 8 cm, then PB equals:
A tangent at point P on a circle with center O makes what angle with the radius OP?
The maximum number of parallel tangents that can be drawn to a circle is:
Sample Questions
From an external point T, tangents TA and TB are drawn to a circle with center O. If ∠ATB = 60°, then ∠AOB equals:
Show answer
120°
Step 1: We have tangents TA and TB from external point T, with ∠ATB = 60°. Step 2: In quadrilateral TAOB, ∠TAO = ∠TBO = 90° (tangent perpendicular to radius). Step 3: Sum of angles in quadrilateral = 360°, so ∠ATB + ∠AOB + ∠TAO + ∠TBO = 360°. Step 4: Therefore, 60° + ∠AOB + 90° + 90° = 360°, which gives ∠AOB = 120°.
If the radius of a circle is 5 cm and the length of a tangent from an external point is 12 cm, then the distance from the external point to the center is:
Show answer
13 cm
Step 1: Let P be the external point, T be the point of contact, and O be the center. Step 2: We have OT = 5 cm (radius) and PT = 12 cm (tangent length). Step 3: Since tangent is perpendicular to radius, triangle POT is a right triangle with right angle at T. Step 4: Using Pythagoras theorem: PO² = PT² + OT² = 12² + 5² = 144 + 25 = 169. Therefore, PO = 13 cm.
A chord of length 16 cm is at a distance of 6 cm from the center of a circle. The radius of the circle is:
Show answer
10 cm
Step 1: Let the chord be AB with length 16 cm, and let M be the midpoint where perpendicular from center O meets AB. Step 2: Since perpendicular from center bisects the chord, AM = MB = 8 cm. Step 3: Given that OM = 6 cm (distance from center to chord). Step 4: In right triangle OMA, using Pythagoras theorem: OA² = OM² + AM² = 6² + 8² = 36 + 64 = 100. Therefore, radius OA = 10 cm.
Two tangents to a circle from an external point make an angle of 80° with each other. The angle between the two radii to the points of contact is:
Show answer
100°
Step 1: Let the external point be P and tangents be PA and PB with ∠APB = 80°. Step 2: In quadrilateral PAOB, ∠PAO = ∠PBO = 90° (tangent perpendicular to radius). Step 3: Sum of angles in quadrilateral = 360°, so ∠APB + ∠AOB + ∠PAO + ∠PBO = 360°. Step 4: Therefore, 80° + ∠AOB + 90° + 90° = 360°, which gives ∠AOB = 100°.
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Sources & Official References
- Karnataka SSLC — kseeb.kar.nic.in
- Dept of Pre-University Education, Karnataka
- National Education Policy 2020 — education.gov.in
Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.
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