A) \[{{x}^{2}}+{{y}^{2}}+4x-6y+19=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x-6y+29=0\]
D) \[{{x}^{2}}+{{y}^{2}}-6x-4y+19=0\]
Correct Answer: B
Solution :
Let, centre of given circle \[{{x}^{2}}+{{y}^{2}}-6x-4y-11=0\] is 0(3, 2). For required circle, P(1, 8) and 0(3, 2) will be the end points of its diameter. \[\therefore \,\,\,\left( x-1 \right)\left( x-3 \right)+\left( y-8 \right)\left( y-2 \right)=0\] [From figure \[{{\operatorname{OP}}^{2}}-P{{A}^{2}}= Radius of\,\,circumcircle\] so P and O are end point of diameter] \[\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]cYou need to login to perform this action.
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