question_answer

If a circle C passing through the point (4, 0) touches the circle \[{{x}^{2}}+{{y}^{2}}+4x-6y=12\] externally at the point \[\left( 1,\,\,-1 \right)\], then the radius of C is- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) 5                                             

B) \[2\sqrt{5}\]

C) 4                     

D)                  \[\sqrt{57}\]

Correct Answer: A

Solution :

Circle C   \[{{S}_{1}}+\lambda {{S}_{2}}\,=\,O\] \[{{x}^{2}}+{{y}^{2}}+4x-6y-12+\lambda ({{(x-1)}^{2}}+{{(y+1)}^{2}})=0\] it passes through (4, 0) \[16+0+16-0-12+\lambda (9+1)=0\] \[\lambda =-2\] \[{{x}^{2}}+{{y}^{2}}+4x-6y-12-2\left( {{x}^{2}}+{{y}^{2}}-2x+2y+2 \right)\] = 0 \[{{x}^{2}}+{{y}^{2}}-8x+10y+16=0\] \[r=\sqrt{6+25-16}\,\,=\,5\]