question_answer

The equation of a circle of radius 5 which lies within the circle \[{{x}^{2}}+{{y}^{2}}+14x+10y-26=0\]and touches it at the point \[(-1,3)\] is

A)  \[{{x}^{2}}+{{y}^{2}}+\text{ }8x+2y+8=0\]

B)  \[{{x}^{2}}+{{y}^{2}}+8x+2y-8=0\]

C)  \[{{x}^{2}}+{{y}^{2}}+8x+2y-14=0\]

D)  None of the above

Correct Answer: B

Solution :

The equation of given circle is \[{{x}^{2}}+{{y}^{2}}+14x+10y-26=0\]                     ?(i) Its centre\[(-g,-f)=(-7-5)\] and radius \[=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]                                 \[=\sqrt{49+25+26}=10\] Also, the required circle is of radius 5 or diameter 10. Therefore, it passes through the centre of the circle (i). Thus, the points \[(-7,-5)\] and \[(-1,3)\] are the end points of the diameter of required circle which is given by                 \[(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0\]                 where    \[({{x}_{1}},{{y}_{1}})=(-7,-5)\] and       \[({{x}_{2}},{{y}_{2}})=(-1,3)\] \[\therefore \]Required circle is given by                 \[(x+7)(x+1)+(y+5)(y-3)=0\]                 \[\Rightarrow \]\[{{x}^{2}}+8x+7+{{y}^{2}}+2y-15=0\]                 \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+8x+2y-8=0\]