Question:medium

The number of common tangents to two circles \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 8x + 12 = 0\) is:

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Externally touching circles (\(d = r_1 + r_2\)) have 3 common tangents.

Updated On: Apr 16, 2026

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The Correct Option is C

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The number of common tangents to two circles \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 8x + 12 = 0\) is:

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The Correct Option is C

Solution and Explanation

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