GSEB Class 9 Maths Solutions Chapter 10 Circles Exercise 10.3

Gujarat Board Textbook Solutions Class 9 Maths Chapter 10 Circles Ex 10.3

Question 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Answer: We can draw pairs of circles in these ways:
(i) With no points in common (they don't touch).

(ii) With one point in common (they touch at a single spot).

(iii) With two points in common (they cross each other at two places).

The highest number of shared points a pair of circles can have is two.
In simple words: When drawing two circles, they can either not touch, touch at one point, or cross at two points. The most points they can share is two.

Exam Tip: Remember that two circles can intersect at most at two points; more than two common points would mean they are the same circle.

Question 2. Suppose you are given a circle. Give the construction to find its centre.
Answer: To find the centre of a given circle, follow these steps:
(i) Choose any three points, \(P\), \(Q\), and \(R\), on the circle's edge.
(ii) Connect point \(P\) to \(Q\), and point \(Q\) to \(R\) with lines.
(iii) Create perpendicular bisectors for both line segment \(PQ\) and line segment \(QR\). These bisectors will meet at a point, let's call it \(O\). This point \(O\) is the circle's exact centre.

In simple words: To find a circle's middle point, pick three spots on its edge, then draw lines connecting them to make two chords. The point where the middle-cutting lines (perpendicular bisectors) of these chords cross is the circle's center.

Exam Tip: The perpendicular bisector of any chord of a circle always passes through the center of the circle. This property is crucial for locating the center.

Question 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Answer:Given: Two circles with centres \(O\) and \(P\) intersecting at \(A\) and \(B\).
Prove: \(OP\) is the perpendicular bisector of \(AB\).
Construction: Join \(OA\), \(OB\), \(PA\) and \(PB\). Let \(OP\) intersect \(AB\) at \(M\).

Proof: Let's consider the triangles \( \triangle OAP \) and \( \triangle OBP \):
\( OA = OB \) (These are the radii of the same circle with center \(O\)).
\( PA = PB \) (These are the radii of the same circle with center \(P\)).
\( OP = OP \) (This side is shared by both triangles).

\( \implies \) Therefore, \( \triangle OAP \cong \triangle OBP \) (By the SSS (Side-Side-Side) congruence rule).

\( \implies \) This means \( \angle AOP = \angle BOP \) (By CPCT - Corresponding Parts of Congruent Triangles).

\( \implies \) So, we can say \( \angle AOM = \angle BOM \) (Let's call this relationship (1)).
Now, let's look at \( \triangle AOM \) and \( \triangle BOM \):
\( OA = OB \) (Again, these are radii of the circle with center \(O\)).
\( \angle AOM = \angle BOM \) (We established this from relationship (1)).
\( OM = OM \) (This side is common to both triangles).

\( \implies \) Therefore, \( \triangle AOM \cong \triangle BOM \) (By the SAS (Side-Angle-Side) congruence rule).

\( \implies \) This means \( AM = BM \) (By CPCT - Corresponding Parts of Congruent Triangles, let's call this (2)).

\( \implies \) Also, \( \angle AMO = \angle BMO \) (By CPCT, let's call this (3)).
However, we know that \( \angle AMO + \angle BMO = 180^\circ \) (Because they form a linear pair on a straight line).
Since \( \angle AMO \) and \( \angle BMO \) are equal (from (3)) and their sum is \( 180^\circ \),

\( \implies \) Each angle must be \( 90^\circ \). So, \( \angle AMO = \angle BMO = 90^\circ \) (Let's call this (4)).

\( \implies \) From (2), we know that \( AM = BM \) (meaning \(M\) bisects \(AB\)).

\( \implies \) From (4), we know that \( OM \) is perpendicular to \( AB \).

\( \implies \) Therefore, \( OM \), which is part of the line \( OP \), is the perpendicular bisector of the common chord \( AB \). This proof comes directly from results (2) and (4).
In simple words: When two circles cross each other, the line that connects their centers cuts the shared line (the common chord) exactly in half and at a perfect right angle. This means the centers always lie on the special line that bisects the common chord perpendicularly.

Exam Tip: This is a fundamental theorem in circle geometry. Remember to clearly state the congruence criteria (SSS, SAS, etc.) and the reasons for each step (Radii of a circle, Common, CPCT, Linear Pair).