CBSE Solutions for Class 10 Maths Chapter 10: Circles || CBSE Class 10 Maths Chapter 10 Circles solutions Ex 10.1

The common point of a tangent to a circle and the circle is called the Point of contact.

A is point of contact here.

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
\( \mathrm{OP} \perp \mathrm{PQ} \)
By applying Pythagoras theorem in \(\triangle \mathrm{OPQ} \),

\(
\mathrm{OP}^{2}+\mathrm{PQ}^{2}=\mathrm{OQ}^{2}\)
\(5^{2}+\mathrm{PQ}^{2}=12^{2}\)
\(\mathrm{PQ}^{2}=144-25\)
\(\mathrm{PQ}=\sqrt{119} \mathrm{~cm}
\)

Steps of Construction:
1) Draw a circle with any radius and center \(\mathrm{O} \).
(2) Choose any point \(\mathrm{ P} \) on the circumference of circle, and draw a line passing through \( \mathrm{P}\), Let's name it CD.
3) Draw a line \( \mathrm{A B} \) parallel to \( \mathrm{ C D} \), such that \( \mathrm{A B} \) intersects the circle at two points \( \mathrm{P} \) and \( \mathrm{A} \). Here, \( \mathrm{A B} \) and \(\mathrm{CD}\) are two parallel lines. \(\mathrm{AB}\) intersects the circle at exactly two points, \(\mathrm{P}\) and \(\mathrm{Q}\). Therefore, line \(\mathrm{AB}\) is the secant of this circle. \(\mathrm{CD}\) intersects the circle at exactly one point, \(\mathrm{R}\), line \(\mathrm{CD}\) is the tangent to the circle.