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

Learn how to find center and radius of circle here.

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.

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