Circle: A circle is a locus of a point that moves in such a way that the distance from that point is always fixed.
Parts of a Circle
The theorem states that “the tangent to the circle at any point is perpendicular to the radius of the circle that passes through the point of contact”.
In the above diagram l is tangent to the circle at point P. O is the center, Then OP is perpendicular to l.
OP ⊥ l.
Example 1: In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then calculate ∠BAT.
Solution:
∠1 = ∠2
∠1 + ∠2 + 100° = 180°
∠1 + ∠1 = 80°
⇒ 2∠1 = 80°
⇒ ∠1 = 40°
∠1 + ∠BAT = 90°
∠BAT = 90° – 40° = 50°
If a point lies inside a circle, any line passing through that point will intersect the circle at two points and is called a secant. Therefore, it is not possible to draw a tangent to a circle that passes through a point inside the circle.
When a point lies on the circle, there is exactly one tangent to a circle that passes through it.
If a point is located outside of a circle, then there exist exactly two tangents that can be drawn to the circle passing through the point.
The lengths of tangents drawn from an external point to a circle are equal.
In the above diagram, two tangents, AP and BP are drawn from external point P. The theorem states that the length of both tangents will be equal.
AP = BP
Example 2: In the given figure, AB and AC are tangents to the circle with centre o such that ∠BAC = 40°. Then calculate ∠BOC.
Solution:
AB and AC are tangents
∴ ∠ABO = ∠ACO = 90°
In ABOC,
∠ABO + ∠ACO + ∠BAC + ∠BOC = 360°
90° + 90° + 40° + ∠BOC = 360°
∠BOC = 360 – 220° = 140°
Example 3: In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then calculate ∠OAB.
Solution:
∠1 = ∠2
∠1 + ∠2 + ∠APB = 180°
∠1 + ∠1 + 60° = 180°
2∠1 = 180° – 60° = 120°
1
∠1 = 60°
∠1 + ∠OAB = 90°
60° +∠OAB = 90°
∠OAB = 90° – 60° = 30°
Example 4: In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then calculate ∠TP.
Solution:
∠1 = ∠2
∠1 + ∠2 + 70° = 180°
∠1 + ∠1 = 180° – 70°
2∠1 = 110° ⇒ ∠1 = 55°
∠1 + ∠TPQ = 90°
55° + ∠TPQ = 90°
⇒ ∠TPQ = 90° – 55° = 35°
Example 5: In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.
Solution:
∠ACB = 90° …[Angle in the semicircle
In ∆ABC,
∠CAB + ∠ACB + ∠CBA = 180°
30 + 90° + ∠CBA = 180°
∠CBA = 180° – 30° – 90° = 60°
∠PCA = ∠CBA …[Angle in the alternate segment
∴ ∠PCA = 60°
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