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lines 5x+12y-10=0 & 5x-12y-40=0 touch a circle of diameter 6 find equatoin of another circle which is concentric with the given circle and c uts intercepts of lenght 8 on these lines

lines 5x+12y-10=0 & 5x-12y-40=0 touch a circle of diameter 6 find equatoin of another circle which is concentric with the given circle and c uts intercepts of lenght 8 on these lines

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1 Answers

Arun
25750 Points
3 years ago
If C be the centre of the circle C 
1
​ 
  then it lies on the bisectors of the given lines which are
13
5x+12y−10
​ 
 =± 
13
5x−12y−40
​ 
 
These give x=5 and y=−5/4. Since centre lies in 1st quadrant therefore y=−5/4 is ruled out. Let the centre be (5,k) then its perpendicular distance from each of lines will be 3,
i.e. p=r.
∴ 
13
25+12kl−10
​ 
 =3 and  
13
25−12k−40
​ 
 =3
∴k=2 or k=− 
12
54
​ 
 =− 
2
9
​ 
 
The value −9/2 of k is ruled out as centre C lies in 1st quadrant ∴C is (5,2).
Now the circle C 
2
​ 
  cuts off intercepts 8 from these lines and if its radius be r, then
r 
2
 =3 
2
 +4 
2
 =25 or r=5
∴ Circle C 
2
​ 
  is (x−5) 
2
 +(y−2) 
2
 =5 
2
 
or x 
2
 +y 
2
 −10x−4y+4=0.

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