Find the number of common tangents to the circles \(x^2 + y^2\) = 1 and \(x^2 + y^2 - 2x - 6y + 6\) = 0.

Solution :

Let \(C_1\) be the center of circle \(x^2 + y^2\) = 1 i.e. \(C_1\) = (0, 0)

And \(C_2\) be the center of circle \(x^2 + y^2 – 2x – 6y + 6\) = 0 i.e. \(C_2\) = (1, 3)

Let \(r_1\) be the radius of first circle and \(r_2\) be the radius of second circle.

Then, \(r_1\) = 1 and \(r_2\) = 2

Now, \(C_1C_2\) = \(\sqrt{9 + 1}\) = \(\sqrt{10}\) and \(r_1 + r_2\) = 3

Since, \(C_1C_2\) > \(r_1 + r_2\)

Hence, there are four common tangents.