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CBSE Class 10 Maths Chapter 10: Circles Notes

📖 Chapter Notes ✏️ NCERT Solutions 📥 PDF Notes

1. Tangent to a Circle

A tangent is a straight line that touches the circle at exactly one single point. The point where the line meets the circle is known as the point of contact.

2. Core Circle Theorems (The Big Two)

These two properties are used to solve every mathematical problem in this chapter:

👉 Theorem 10.1 (Radius-Tangent Perpendicularity):
The tangent at any point of a circle is strictly perpendicular ($90^\circ$) to the radius through the point of contact.

👉 Theorem 10.2 (Equal Tangent Lengths):
The lengths of tangents drawn from an external point to a circle are completely equal. (If point $P$ is outside, and tangents touch at $A$ and $B$, then $PA = PB$).
CBSE Board Exam Classic Formula

Pythagoras Connection: Because the radius and tangent form a perfect $90^\circ$ right angle, you will almost always use the Pythagoras Theorem ($H^2 = P^2 + B^2$) to calculate missing lengths linking the center of the circle, the external point, and the tangent point!


✏️ Complete NCERT Solutions Class 10 Circles

Exercise 10.1
Q1. How many tangents can a circle have?
Answer: A circle is made up of an infinite number of points placed together. Since a unique tangent can be drawn at every single point, a circle can have infinitely many tangents.
Exercise 10.2
Q2. From a point $Q$, the length of the tangent to a circle is $24\text{ cm}$ and the distance of $Q$ from the center is $25\text{ cm}$. Find the radius of the circle.
Step 1: Visualize the right-angled triangle ($\Delta OPQ$)
Let $O$ be the center, $P$ be the point of contact, so $OP$ is the radius ($r$).
Given: Tangent $PQ = 24\text{ cm}$ and Hypotenuse from center $OQ = 25\text{ cm}$.
Step 2: Apply the Pythagoras Theorem ($\angle OPQ = 90^\circ$)
$OQ^2 = OP^2 + PQ^2 \implies 25^2 = r^2 + 24^2$
Step 3: Solve for radius ($r$)
$625 = r^2 + 576$
$r^2 = 625 - 576 = 49$
$r = \sqrt{49} = \mathbf{7\text{ cm}}$
Final Answer: The radius of the circle is $7\text{ cm}$.