question_answer

The centre of the circle passing through the point (0,1) and touching the curve y = x2 at point (2, 4) is

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 \[\left( -\frac{16}{5},\frac{53}{10} \right).\]