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

To find the number of common tangents between the two given circles, we first need to analyze their equations and geometric properties:

The equation of the first circle is \(x^2 + y^2 = 4\). This equation represents a circle centered at \((0, 0)\) with a radius of \(2\).

The equation of the second circle is \(x^2 + y^2 - 8x + 12 = 0\). We can simplify this equation by completing the square:

• Rewriting: \((x^2 - 8x) + y^2 = -12\).
• Completing the square for the \(x\)-terms: \(x^2 - 8x = (x - 4)^2 - 16\).
• Substituting back, we get: \((x - 4)^2 + y^2 = 4\).

Next, determine the distance between the centers of the two circles:

The distance \(d\) is calculated as:

1. \(d = \sqrt{(4 - 0)^2 + (0 - 0)^2} = \sqrt{16} = 4\).

To find the number of common tangents:

• Since the distance between the centers \(d = 4\) is equal to the sum of the radii \(2 + 2 = 4\), the circles are externally tangent to each other.
• For two externally tangent circles, there are three common tangents: two external tangents and one internal tangent.

Based on this information, the number of common tangents to the two circles is 3. Thus, the correct answer is:

3