A) \[\left( -\frac{16}{5},\frac{27}{10} \right)\]
B) \[\left( -\frac{16}{7},\frac{53}{10} \right)\]
C) \[\left( -\frac{16}{5},\frac{53}{10} \right)\]
D) None of these
Correct Answer: C
Solution :
Let the centre be O(h, k). So, \[O{{A}^{2}}=O{{B}^{2}}\]and (slope of OA) (slope of tangent at A) = -1 \[\Rightarrow \]\[{{(h-2)}^{2}}+{{(k-4)}^{2}}={{h}^{2}}+{{(k-1)}^{2}}\] \[\Rightarrow \]\[4h+6k-19=0\] ?(i) Also, slope of\[OA=\frac{k-4}{h-2}\]and equation of tangent at (2, 4) to y = x2 is y + 4 = 2x .2, its slope is 4. \[\therefore \] \[\frac{k-4}{h-2}.4=-1\]\[(\because {{m}_{1}}{{m}_{2}}=-1)\] \[\Rightarrow 4k-16=-h+2\] \[\Rightarrow h+4k=18\] ?(ii) On solving Eqs. (i) and (ii), we get \[h=-\frac{16}{5}\]and\[k=\frac{53}{10}\] \[\therefore \]Coordinates of centre are \[\left( -\frac{16}{5},\frac{53}{10} \right).\]You need to login to perform this action.
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