question_answer

The equation of circle which touches the axes and the line \[\frac{x}{3}+\frac{y}{4}=1\] and whose centre lies in the first quadrant is\[{{x}^{2}}+{{y}^{2}}-2cx-2cy+{{c}^{2}}=0\]. Then, \[c\] is equal to

A) \[1\]                                     

B) \[2\]

C) \[3\]                                     

D) .\[6\]

Correct Answer: A

Solution :

In centre of triangle \[OAB\] is \[\left( \frac{0\times 5+3\times 4+0\times 3}{3+4+5},\,\,\frac{0\times 5+0\times 4+4\times 3}{3+4+5} \right)\]                 \[=(1,\,\,1)\] \[\therefore \]Equation of circle which touches both coordinates is                 \[{{(x-1)}^{2}}+{{(y-1)}^{2}}=1\] \[\Rightarrow \]               \[{{x}^{2}}-2x+{{y}^{2}}-2y+1=0\] On comparing with                 \[{{x}^{2}}+{{y}^{2}}-2cx-2cy+{{c}^{2}}=0\], we get                 \[c=1\]