question_answer

The circle\[{{x}^{2}}+{{y}^{2}}-2cx-2cy+{{c}^{2}}=0\]touches both the axes and the line\[\frac{x}{3}+\frac{y}{4}=1\]. Its centre lies in the first quadrant, then the value of c is

A)  1                 

B)  2

C)  3                 

D)  6

Correct Answer: A

Solution :

 Since, the circle touches both the axes and the line. \[\therefore \]Centre of circle = Incentre of\[\Delta OAB\] \[=\left( \frac{5\times 0+4\times 3+3\times 0}{4+3+5},\frac{0\times 5+4\times 0+3\times 4}{4+3+5} \right)\] \[=(1,1)\] \[\therefore \]Equation of circle touches both axes is \[{{(x-1)}^{2}}+{{(y-1)}^{2}}=1\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\] On comparing with\[{{x}^{2}}+{{y}^{2}}-2cx-2cy+{{c}^{2}}=0,\]we get \[c=1\]