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A common tangent to the circles x^(2)+y^...

A common tangent to the circles `x^(2)+y^(2)=4` and `(x-3)^(2)+y^(2)=1`, is

A

x=4

B

y=2

C

`x+sqrt(3)y=4`

D

`x+2sqrt(2)y=6`

Text Solution

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The correct Answer is:
To find the common tangent to the circles given by the equations \(x^2 + y^2 = 4\) and \((x - 3)^2 + y^2 = 1\), we will follow these steps: ### Step 1: Identify the centers and radii of the circles 1. The first circle \(x^2 + y^2 = 4\) has: - Center \(C_1(0, 0)\) - Radius \(r_1 = \sqrt{4} = 2\) 2. The second circle \((x - 3)^2 + y^2 = 1\) has: - Center \(C_2(3, 0)\) - Radius \(r_2 = \sqrt{1} = 1\)
To find the common tangent to the circles given by the equations \(x^2 + y^2 = 4\) and \((x - 3)^2 + y^2 = 1\), we will follow these steps: ### Step 1: Identify the centers and radii of the circles 1. The first circle \(x^2 + y^2 = 4\) has: - Center \(C_1(0, 0)\) - Radius \(r_1 = \sqrt{4} = 2\) 2. The second circle \((x - 3)^2 + y^2 = 1\) has: ...
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Knowledge Check

  • The length of the transversal common tangent to the circle x^(2)+y^(2)=1 and (x-t)^(2)+y^(2)=1 is sqrt(21) , then t=

    A
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    B
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    D
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    B
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    C
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    D
    `x^(2)-y^(2)=1`

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    A
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    C
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