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`        sir pls solve ths qutnthe circle x2+y2-6x-10y+c=0 does not touch or intersect the co-ordinate axes and the point(1,4) is inside the circle.find the set of values of c.`
8 years ago

147 Points
```										Dear rohan
x2+y2-6x-10y+c=0
(x-3)2 +(y-5)2 = 34-c
circle does not intersect the co ordinate axis so √(34-c) <3
or 34-c<9
or  c >25

and 1,4 lie inside the circle so redius of circle must be greater then distance between the point (1,4) and (3,5)
√(34-c)> √(3-1)2 +(5-4)2
34-c  >√5
c < 34- √5
so   25 < c  < 34- √5

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```
8 years ago
sid
22 Points
```										But sir ,answer that is mentioned here in book is (25,29).….......…....................///.........................
```
one year ago
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