RD Sharma Solutions Chapter 10 Circles Exercise 10.1 Class 10 Maths
Chapter Name | RD Sharma Chapter 10 Circles |
Book Name | RD Sharma Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 10.1 Solutions
1. Fill in the blanks :
(i) The common point of a tangent and the circle is called ……
(ii) A circle may have …… parallel tangents.
(iii) A tangent to a circle intersects it in ……. point(s).
(iv) A line intersecting a circle in two points is called a ……
(v) The angle between tangent at a point on a circle and the radius through the point is ……
Solution
(i) The common point of a tangent and the circle is called the point of contact.
(ii) A circle may have two parallel tangents.
(iii) A tangents to a circle intersects it in one point.
(iv) A line intersecting a circle in two points is called a secant.
(v) The angle between tangent at a point, on a circle and the radius through the point is 90°.
2. How many tangents can a circle have ?
Solution
A circle can have infinitely many tangents as there can many point lie on the circumference of the circle.
3. O is the centre of a circle of radius 8 cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB.
Solution
Radius OA = 8 cm, ST is the tangent to the circle at A and AB = 15 cm
OA ⊥ tangent TS
In right ∆OAB,
OB² = OA² + AB² (Pythagoras Theorem)
⇒ OB² = (8)² + (15)²
⇒ OB² = 64 + 225
⇒ OB² = 289
⇒ OB² = (17)²
⇒ OB = 17 cm
4. If the tangent at a point P to a circle with centre O cuts a line through O at Q such that PQ = 24 cm and OQ = 25 cm. Find the radius of the circle.
Solution
OP is the radius and TS is the tangent to the circle at P
OQ is a line
OP ⊥ tangent TS
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
⇒ (25)² = OP² + (24)²
⇒ 625 = OP² + 576
⇒ OP² = 625 – 576 = 49
⇒ OP² = (7)²
⇒ OP = 7 cm
Hence, radius of the circle is 7 cm.