NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.2
Question 5

Question. 5

In Fig. 9.13, \(AB\) and \(CD\) are common tangents to two circles of unequal radii. Prove that \(AB=CD\).

Fig. 9.13 (update src later)

Answer:

\(AB=CD\)

Detailed Answer with Explanation:

Step 1: Let the two common tangents \(AB\) and \(CD\) meet at a point \(P\) outside the circles.

Step 2: From geometry, we know that the lengths of tangents drawn from the same external point to a circle are equal.

Step 3: Apply this property to the left circle:

  • \(PA\) and \(PC\) are tangents from point \(P\).
  • Therefore, \(PA = PC\).

Step 4: Apply the same property to the right circle:

  • \(PB\) and \(PD\) are tangents from point \(P\).
  • Therefore, \(PB = PD\).

Step 5: Look at line \(PAB\):

  • \(AB = PA - PB\).

Step 6: Now look at line \(PCD\):

  • \(CD = PC - PD\).

Step 7: Since \(PA = PC\) and \(PB = PD\), subtracting gives:

\[ AB = PA - PB = PC - PD = CD \]

Final Result: Hence, we have proved that \(AB = CD\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 9: Circles – Exercise 9.2 | Detailed Answers