A) \[{{x}^{2}}+{{y}^{2}}-2x+4y+1=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x-4y+1=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x+4y+1=0\]
D) \[{{x}^{2}}+{{y}^{2}}+4x+2y+4=0\]
Correct Answer: B
Solution :
Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2. Hence the equation is \[{{(x-1)}^{2}}+{{(y-2)}^{2}}={{2}^{2}}\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-4y+1=0\]. Trick : The only circle is \[{{x}^{2}}+{{y}^{2}}-2x-4y+1=0\], whose centre is (1, 2).
You need to login to perform this action.
You will be redirected in
3 sec
