Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

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

Answer 1

The Correct Option is C

Concept: Number of common tangents depends on the distance between centers relative to radii.

Step 1: Find centers and radii. Circle 1: \(x^2 + y^2 = 4\) → center \(C_1 = (0,0)\), radius \(r_1 = 2\). Circle 2: \(x^2 + y^2 - 8x + 12 = 0\) → \((x - 4)^2 + y^2 = 4\) → center \(C_2 = (4,0)\), radius \(r_2 = 2\).

Step 2: Distance between centers. \[ d = \sqrt{(4-0)^2 + (0-0)^2} = 4 \]

Step 3: Compare \(d\) with \(r_1 + r_2\). \[ r_1 + r_2 = 2 + 2 = 4 \Rightarrow d = r_1 + r_2 \] Circles touch externally. Number of common tangents = 3 (2 direct + 1 transverse).