Circle 2 Question 11
12. The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6 x-8 y=24$ is
$(1998,2 M)$
(a) 0
(b) 1
(c) 3
(d) 4
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Answer:
Correct Answer: 12. (b)
Solution:
- Given, $x^{2}+y^{2}=4$
Centre $\equiv C _1 \equiv(0,0)$ and $R _1=2$
Again, $x^{2}+y^{2}-6 x-8 y-24=0$, then $C _2 \equiv(3,4)$
and $ R _2=7 $
Again, $ C _1 C _2=5=R _2-R _1 $
Therefore, the given circles touch internally such that, they can have just one common tangent at the point of contact.
