NCERT Class 10 Mathematics — Circles, Exercise 10.1. All 4 questions solved.

Exercise 10.1 is the vocabulary exercise of the chapter, with one short calculation. Everything in it rests on three ideas from the first half of the chapter:

• A secant is a line that meets a circle in two points. A tangent is a line that meets it in exactly one point, called the point of contact.
• At every point of a circle there is one and only one tangent.
• Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Key insight. A tangent is best thought of as the limiting position of a secant whose two meeting points have slid together into one. That picture explains the whole exercise: why a tangent touches in exactly one point, why there is one tangent at each point of the circle, and why the radius to that point meets it at a right angle — which is the only fact question 3 needs.

Question 1

How many tangents can a circle have?

Solution. Each point of the circle has exactly one tangent through it, and a circle has infinitely many points. Different points give different tangents, because a tangent touches the circle at only one point. So the tangents are as many as the points of the circle.

Question 2

Fill in the blanks:

(i) A tangent to a circle intersects it in ______ point (s). (ii) A line intersecting a circle in two points is called a ______. (iii) A circle can have ______ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called ______.

Solution.

(i) One. This is the definition of a tangent: the two meeting points of a secant have merged into a single point.

(ii) Secant. A line through two points of the circle. The part of it inside the circle, between the two points, is a chord; the secant is the whole line.

(iii) Two. Fix a direction. A tangent in that direction must be perpendicular to the radius at its point of contact, so its point of contact lies on the diameter perpendicular to that direction. A diameter has only two ends, so there can be at most two tangents parallel to each other — one at each end. (There are infinitely many such pairs, one pair for each direction, but never three mutually parallel tangents.)

(iv) Point of contact. The single point that a tangent and the circle share.

Question 3

A tangent $\mathrm{PQ}$ at a point $\mathrm{P}$ of a circle of radius $5$ cm meets a line through the centre $\mathrm{O}$ at a point $\mathrm{Q}$ so that $\mathrm{OQ} = 12$ cm. Length $\mathrm{PQ}$ is:

Solution. $\mathrm{OP}$ is the radius to the point of contact of the tangent $\mathrm{PQ}$, so by Theorem 10.1, $\mathrm{OP} \perp \mathrm{PQ}$. Triangle $\mathrm{OPQ}$ is therefore right-angled at $\mathrm{P}$, and its hypotenuse is $\mathrm{OQ}$, the side opposite the right angle. By Pythagoras:

$$\mathrm{PQ}^2 = \mathrm{OQ}^2 – \mathrm{OP}^2 = 12^2 – 5^2 = 144 – 25 = 119$$

$$\mathrm{PQ} = \sqrt{119}\ \text{cm} \approx 10.9\ \text{cm}$$

As a sanity check, $\mathrm{PQ}$ must be shorter than the hypotenuse $12$ cm, which rules out (A) and (B) straight away.

Question 4

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution. The tangent is the only part that needs care, and Theorem 10.1 tells us how to draw it: a tangent is perpendicular to the radius at its point of contact. So if we want a tangent parallel to the given line $l$, its point of contact must be at the end of the diameter that is perpendicular to $l$.

Construction.

1. Draw a circle with centre $\mathrm{O}$ and any radius, and draw the given line $l$.
2. Through $\mathrm{O}$, draw a line perpendicular to $l$ (dashed in the figure). It meets the circle at two points; call one of them $\mathrm{P}$.
3. At $\mathrm{P}$, draw the line perpendicular to $\mathrm{OP}$. This is the tangent at $\mathrm{P}$.
4. Take any point of the dashed diameter strictly between its two ends, and through it draw a line parallel to $l$. It cuts the circle at two points $\mathrm{A}$ and $\mathrm{B}$, so it is a secant.

Why it works. The tangent at $\mathrm{P}$ and the line $l$ are both perpendicular to the same line (the dashed diameter), so they are parallel. The line through $\mathrm{A}$ and $\mathrm{B}$ was drawn parallel to $l$, and its distance from $\mathrm{O}$ is less than the radius, which is exactly the condition for a line to cut a circle in two points.

Common mistakes

• Question 1, answering “two”. Two is the number of tangents from a point outside the circle, which is a different question. The circle as a whole has one tangent at every one of its infinitely many points.
• Question 2(ii), writing “chord”. A chord is the segment joining two points of the circle; the line through them is the secant. The blank asks for a line.
• Question 2(iii), writing “infinitely many”. A circle has infinitely many pairs of parallel tangents, but any family of tangents all parallel to one another has at most two members — one at each end of a diameter.
• Question 3, putting the right angle at O. Students who do this compute $\sqrt{12^2 + 5^2} = 13$ and pick (B). The right angle is always at the point of contact, between the radius and the tangent, so $\mathrm{OQ}$ is the hypotenuse.
• Question 3, “simplifying” $\sqrt{119}$. $119 = 7 \times 17$ has no square factor, so $\sqrt{119}$ is already in its simplest form. Option (C), $8.5$ cm, is the average of $12$ and $5$ and is a trap for guessing.
• Question 4, drawing the tangent by eye. A line that merely looks as if it grazes the circle is not a construction. Draw the diameter perpendicular to $l$ first; the tangent is then fixed as the perpendicular at its end.

Practise next

• Exercise 10.2 — the second theorem of the chapter, that the two tangents from an external point are equal, and the proofs built from it.
• Chapter 11, Exercise 11.1 — sectors and segments, where the triangle made by two radii and a chord comes back in almost every question.