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. 5
• D. 3

Answer 1

Answer: (D)

Since, centre and radius of a circle $ x^{2}+y^{2}=4 $ are
$ C_{1}(0, 0) $ and $ 2 $ respectively
and centre and radius of another circle
$ x^{2}+y^{2}-8x+12=0 $ are
$ C_{2}(4, 0) $ and $ 2 $ respectively.
Now, $ C_{1}C_{2}=\sqrt{(4-0)^{2}+0=4}$
and $ r_{1}+r_{2}=2+2=4 $
$ \because C_{1}C_{2}=r_{1}+r_{2} $
$ \therefore $ Two circles touch each other externally,
so the number of common tangents is $ 3 $.
Note: (i) If $ C_{1}C_{2}=r_{1}-r_{2} $ , one tangents are possible.
If $ C_{1}C_{2} > r_{1}+r_{2} $ , four tangents are possible.