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Question

There are two circles whose equations are S1=x2+y2=49 and S2=x2+y2+16x+12y+p2=0,   pZ. If these circles have exactly two common tangents, then the sum of all the possible values of p is 


Solution

For circle S2=x2+y2+16x+12y+p2=0, the centre is (8,6) and radius is (8)2+(6)2p2
Then (8)2+(6)2p2>0
10p10
The circles should intersect each other at two points to have exactly two common tangents. So,
r1+r2>d>|r2r1|
d=(80)2+(60)2=10
r1+r2>d7+100p2>10
100p2>3
100p2>9
91p91
Also d>|r2r1|
10>7100p2    or
10>100p27
if 10>7100p2
100p2>3 which is always true
If  10>100p27
p2>189 which will be also always true
so 91p91
​​​​Now pZ so p=9,8,7,.......8,9
sum of all values p=0

 

Mathematics

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