Table of Contents
Exercise 9.1 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
Fill in the blanks:
(i) A tangent to a circle touches it in one point(s).
Explanation: By definition, a tangent touches a circle at exactly one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) Number of tangents can be drawn to a circle parallel to the given tangent is one.
Explanation: For any given tangent, there exists exactly one other tangent parallel to it.
(iv) The common point of a tangent to a circle and the circle is called point of contact.
(v) We can draw infinite tangents to a given circle.
Explanation: There are infinitely many points on a circle, and at each point there’s a unique tangent.
(vi) A circle can have two parallel tangents at the most.
Explanation: A circle can have exactly two parallel tangents – one on each side.
Problem 2
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that \( \text{OQ} = 13 \, \text{cm} \). Find length of PQ.
Solution:
Given: OP = radius = 5 cm, OQ = 13 cm
Since PQ is tangent, OP ⊥ PQ (radius perpendicular to tangent at point of contact)
In right triangle OPQ:
\( OP^2 + PQ^2 = OQ^2 \)
\( 5^2 + PQ^2 = 13^2 \)
\( 25 + PQ^2 = 169 \)
\( PQ^2 = 144 \)
\( PQ = 12 \, \text{cm} \)
Problem 3
Draw a circle and two lines parallel to a given line drawn outside the circle such that one is a tangent and the other, a secant to the circle.
Solution:
Construction Steps:
- Draw a circle with center O and any radius
- Draw a line l outside the circle (not intersecting the circle)
- Draw perpendicular from O to line l, meeting at point P
- With OP as distance, draw line m parallel to l – this will be tangent (touches at one point)
- Draw another line n parallel to l at distance less than OP – this will be secant (intersects at two points)
Problem 4
Calculate the length of tangent from a point 15 cm away from the centre of a circle of radius 9 cm.
Solution:
Given: Distance from center (d) = 15 cm, Radius (r) = 9 cm
Length of tangent (l) from external point is given by:
\( l = \sqrt{d^2 – r^2} = \sqrt{15^2 – 9^2} = \sqrt{225 – 81} = \sqrt{144} = 12 \, \text{cm} \)
Problem 5
Prove that the tangents to a circle at the end points of a diameter are parallel.
Solution:
Let AB be diameter of circle with center O.
Let PA be tangent at A and QB be tangent at B.
Property: Tangent is perpendicular to radius at point of contact.
Thus, PA ⊥ OA and QB ⊥ OB
But OA and OB lie on same line AB (diameter)
Therefore, PA ⊥ AB and QB ⊥ AB
If two lines are both perpendicular to the same line, they are parallel to each other.
Hence, PA ∥ QB
