Home
Class 12
MATHS
The number of common tangents to the cir...

The number of common tangents to the circles `x^(2) + y^(2) = 4 and x^(2)+y^(2)-6x-8y=24` is

A

0

B

1

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To find the number of common tangents to the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 6x - 8y = 24\), we will follow these steps: ### Step 1: Identify the centers and radii of the circles 1. **First Circle**: The equation is \(x^2 + y^2 = 4\). - This can be rewritten as \((x - 0)^2 + (y - 0)^2 = 2^2\). - Center: \(C_1(0, 0)\) - Radius: \(r_1 = 2\) 2. **Second Circle**: The equation is \(x^2 + y^2 - 6x - 8y = 24\). - Rearranging gives \(x^2 - 6x + y^2 - 8y = 24\). - Completing the square: - For \(x\): \(x^2 - 6x = (x - 3)^2 - 9\) - For \(y\): \(y^2 - 8y = (y - 4)^2 - 16\) - Thus, the equation becomes \((x - 3)^2 + (y - 4)^2 - 25 = 0\) or \((x - 3)^2 + (y - 4)^2 = 25\). - Center: \(C_2(3, 4)\) - Radius: \(r_2 = 5\) ### Step 2: Calculate the distance between the centers of the circles - The distance \(d\) between the centers \(C_1(0, 0)\) and \(C_2(3, 4)\) is calculated using the distance formula: \[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] ### Step 3: Determine the relationship between the distance and the radii - The relationship between the distance \(d\), the radii \(r_1\) and \(r_2\) determines the number of common tangents: - If \(d > r_1 + r_2\), there are 4 common tangents. - If \(d = r_1 + r_2\), there are 3 common tangents (the circles touch externally). - If \(|r_1 - r_2| < d < r_1 + r_2\), there are 2 common tangents. - If \(d = |r_1 - r_2|\), there is 1 common tangent (the circles touch internally). - If \(d < |r_1 - r_2|\), there are 0 common tangents (the circles are one inside the other without touching). ### Step 4: Evaluate the conditions - Here, we have: - \(r_1 = 2\) - \(r_2 = 5\) - \(d = 5\) - Now, calculate \(r_1 + r_2\): \[ r_1 + r_2 = 2 + 5 = 7 \] - Since \(d = 5 < 7\), we check \(|r_1 - r_2|\): \[ |r_1 - r_2| = |2 - 5| = 3 \] - Since \(d = 5 > 3\), we conclude that \(d\) is between \(|r_1 - r_2|\) and \(r_1 + r_2\). ### Conclusion Thus, there are **2 common tangents** to the circles. ---

To find the number of common tangents to the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 6x - 8y = 24\), we will follow these steps: ### Step 1: Identify the centers and radii of the circles 1. **First Circle**: The equation is \(x^2 + y^2 = 4\). - This can be rewritten as \((x - 0)^2 + (y - 0)^2 = 2^2\). - Center: \(C_1(0, 0)\) - Radius: \(r_1 = 2\) ...
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    IIT JEE PREVIOUS YEAR|Exercise Topic 2 Objective Question II|2 Videos
  • CIRCLE

    IIT JEE PREVIOUS YEAR|Exercise Topic 2 Passage Based Problems|3 Videos
  • CIRCLE

    IIT JEE PREVIOUS YEAR|Exercise Topic 1 Paragraph Based Questions|2 Videos
  • BINOMIAL THEOREM

    IIT JEE PREVIOUS YEAR|Exercise Topic 2 Properties of Binomial Coefficent Objective Questions I (Only one correct option) (Analytical & Descriptive Questions )|8 Videos
  • COMPLEX NUMBERS

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 5 DE-MOIVRES THEOREM,CUBE ROOTS AND nth ROOTS OF UNITY (INTEGER ANSWER TYPE QUESTION)|1 Videos

Similar Questions

Explore conceptually related problems

Statement 1 : The number of common tangents to the circles x^(2) + y^(2) =4 and x^(2) + y^(2) -6x - 6y = 24 is 3. Statement 2 : If two circles touch each other externally thenit has two direct common tangents and one indirect common tangent.

The number of common tangents to the circles x^(2)+y^(2)-x=0 and x^(2)+y^(2)+x=0 are

The number of common tangents to the circles x^(2)+y^(2)-y=0and x^(2)+y^(2)+y=0 is

The number of common tangents of the circles x^(2) +y^(2) =16 and x^(2) +y^(2) -2y = 0 is :

The number of common tangents to the circle x^(2)+y^(2)-2x-4y-4=0 and x^(2)+y^(2)+4x+8y-5=0 is _________.

The number of common tangents of the circles x^(2) + y^(2) – 2x – 1 = 0 and x^(2) + y^(2) – 2y – 7 = 0