question_answer

Find the centre of a circle passing through the points \[(6,-6),\]\[(3,-7)\]and \[(3,3).\]

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

\[P\,\,(6-6),\]\[Q\,\,(3,-7)\]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

\[3x-2-7=0\]

\[\Rightarrow \]   \[3x=9\]

\[\therefore \]      \[x=3\]