A) \[(3,-2)\]
B) \[(4,5)\]
C) \[(-3,-2)\]
D) \[(-3,2)\]
Correct Answer: A
Solution :
Let C(x, y) be the centre of the circle passing through the points |
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\[P\,\,(6-6),\]\[Q\,\,(3,-7)\]and \[R\,\,(3,3)\] |
Then, \[PC=QC=CR\] (Radius of circle) |
Now, \[PC=QC\] |
\[\Rightarrow \] \[P{{C}^{2}}=Q{{C}^{2}}\] |
\[\Rightarrow \]\[{{(x-6)}^{2}}+{{(y+6)}^{2}}={{(x-3)}^{2}}+{{(y+7)}^{2}}\][\[\because \]Distance\[=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]] |
\[\Rightarrow \]\[{{x}^{2}}-12x+36+{{y}^{2}}+12y+36\]\[={{x}^{2}}-6x+9+{{y}^{2}}+14y+49\] |
\[\Rightarrow \]\[-12x+6x+12y-14y+72-58=0\] |
\[\Rightarrow \] \[-\,\,6x-2y+14=0\] |
\[3x+y-7=0\] ?(i) |
and \[QC=CR\] |
\[\Rightarrow \] \[Q{{C}^{2}}=C{{R}^{2}}\] |
\[\Rightarrow \] \[{{(x-3)}^{2}}+{{(y+7)}^{2}}={{(x-3)}^{2}}+{{(y-3)}^{2}}\] |
\[\Rightarrow \]\[{{x}^{2}}-6x+9+{{y}^{2}}+14y+49\]\[={{x}^{2}}-6x+9+{{y}^{2}}-6y+9\] |
\[\Rightarrow \]\[-\,\,6x+6x+14y+6y+58-18=0\] |
\[\Rightarrow \] \[20y+40=0\] |
\[\Rightarrow \] \[y=-\frac{40}{20}=-2\] ...(ii) |
Putting \[y=-2\] in Eq. (i), we get |
\[3x-2-7=0\] |
\[\Rightarrow \] \[3x=9\] |
\[\therefore \] \[x=3\] |
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