A) \[{{x}^{2}}+{{y}^{2}}-2x+4y-20=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x-4y-20=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x-4y-20=0\]
D) \[{{x}^{2}}+{{y}^{2}}+2x+4y-20=0\]
Correct Answer: A
Solution :
The point of intersection of \[3x+y-14=0\] and \[2x+5y-18=0\] are \[x=\frac{-18+70}{15-2},\ y=\frac{-28+54}{13}\Rightarrow x=4,\ y=2\] i.e., point is (4, 2). Therefore radius is \[\sqrt{(9)+(16)}=5\] and equation is \[{{x}^{2}}+{{y}^{2}}-2x+4y-20=0\]. Trick : The only circle is \[{{x}^{2}}+{{y}^{2}}-2x+4y-20=0\], whose centre is (1, ?2).
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