Question
In the given figure, the equation of the larger circle is x^ 2 +y^ 2 +4 y-5=0 and the distance between centres is 4 . Then the equation of smaller circle is
BITSAT
Mathematics
Circle
2020
BITSAT 2020
BITSAT Mathematics Question (2020) — Solution
Options
- A. (x- 7 )^ 2 +(y-1)^ 2 =1
- B. (x+ 7 )^ 2 +(y-1)^ 2 =1
- C. x^ 2 +y^ 2 =2 7 x+2 y
- D. None of these
Answer
A. (x- 7 )^ 2 +(y-1)^ 2 =1
Step-by-step solution
We have x^ 2 +y^ 2 +4 y-5=0. Its centre is C _ 1 (0,-2) r _ 1 = 4+5 =3. Let C _ 2 ( ~h , k ) be the centre of the smaller circle and its radius r _ 2 . Then C _ 1 C _ 2 =4 h ^ 2 +( k +2)^ 2 =3+ r _ 2 =4 r _ 2 =1 But k = r _ 2 =1 [it touches x -axis From eq (1), 4= h ^ 2 +(1+2)^ 2 16= h ^ 2 +9 h ^ 2 =7 h = 7 Since h >0 h = 7 Hence, required circle is (x- 7 )^ 2 +(y-1)^ 2 =1
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