BITSAT2020MathematicsCircleActual
In the given figure, the equation of the larger circle is x²+y²+4 y-5=0 and the distance between centres is 4 . Then the equation of smaller circle is
Options
- A(x- 7 )²+(y-1)²=1
- B(x+ 7 )²+(y-1)²=1
- Cx²+y²=2 7 x+2 y
- DNone of these
Correct answer
A. (x- 7 )²+(y-1)²=1
Step-by-step solution
We have x²+y²+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