GSEB Class 10 Maths Solutions Chapter 10 Circles Exercise 10.1

Question 1. How many tangents can a circle have?
Answer: A circle can have infinitely many tangents. This is because there are an endless number of points on the circle, and each individual point can have its own tangent line.
In simple words: A circle has countless points, and you can draw a tangent line at every single one of them. So, a circle can have an unlimited number of tangents.

Exam Tip: Remember that a tangent touches the circle at exactly one point. Since a circle has infinite points, it must have infinite tangents.

Question 2. Fill in the blanks:
1. A tangent to a circle intersects it in ......... point(s).
2. A line intersecting a circle in two points is called a ...............
3. A circle may have ......... parallel tangents at the most.
4. The common point of a tangent to a circle and the circle is called ...............
Answer:
1. one
2. secant
3. two
4. point of contact
In simple words: A tangent touches a circle at just one spot. A line that cuts through a circle at two spots is known as a secant. A circle can have a maximum of two parallel tangents, one on each side. The exact point where a tangent meets the circle is called the point of contact.

Exam Tip: Understand the precise definitions of tangent, secant, and point of contact, as these terms are fundamental in geometry.

Question 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13cm
(c) 8.5 cm
(d) 19 cm
Answer: As shown in the figure, OP is the radius of the circle, and PQ is the tangent at point P. We know that the radius and tangent are always perpendicular at the point of contact.
\( \implies \angle OPQ = 90^\circ \) (Theorem 10.1)
Therefore, \( \triangle OPQ \) forms a right-angled triangle.
Using the Pythagoras theorem:
\( OP^2 + PQ^2 = OQ^2 \)
Given \( OP = 5 \) cm and \( OQ = 12 \) cm.
\( 5^2 + PQ^2 = 12^2 \)
\( 25 + PQ^2 = 144 \)
\( PQ^2 = 144 - 25 \)
\( PQ^2 = 119 \)
\( PQ = \sqrt{119} \) cm.
This value is approximately 10.9 cm, which does not match any of the provided options exactly.
In simple words: The radius and tangent meet at a right angle. So, we can use Pythagoras' theorem with the given radius (5 cm) and the distance from the center to the external point (12 cm). We calculate the length of the tangent, which is the square root of 119 centimeters.

Exam Tip: Always draw a clear diagram for geometry problems. Remember that the tangent is always perpendicular to the radius at the point of contact, forming a right-angled triangle, allowing you to use the Pythagorean theorem.

Question 4. Draw a circle and two lines parallel to a given line such that one is tangent and other, a secant to the circle.
Answer: Here are the steps to draw a circle and two lines, one a tangent and one a secant, both parallel to a specified line:
1. Begin by drawing a circle with its center labeled O.
2. Next, draw a line AB that passes directly through the center O. This line serves as your reference.
3. Draw a perpendicular bisector CD of line AB. This line passes through O and is perpendicular to AB.
4. From point D, draw a line segment, say DF, such that \( \angle ODF = 90^\circ \). The line EF will be a tangent to the circle.
5. Take a point G on line segment OD. From G, draw a line segment, say GY, such that \( \angle OGY = 90^\circ \). The line XY will be a secant to the circle.
Thus, the line EF is a tangent, and the line XY is a secant, and both are parallel to line AB.
In simple words: First, draw a circle. Then, draw a straight line (the "given line"). Now, draw another line that only touches the circle at one point (a tangent) and is parallel to your first line. After that, draw a third line that cuts through the circle at two points (a secant), and make sure it's also parallel to your first line.

Exam Tip: For construction problems, always list the steps clearly and label your diagram accurately. Pay attention to properties like perpendicularity and parallelism for lines and circles.