• # question_answer The centre of the circle passing through the point (0, 1) and touching the curve $y={{x}^{2}}$at (2, 4) is       [IIT 1983] A)            $\left( \frac{-16}{5},\ \frac{27}{10} \right)$                              B)            $\left( \frac{-16}{7},\ \frac{5}{10} \right)$ C)            $\left( \frac{-16}{5},\ \frac{53}{10} \right)$                              D)            None of these

Solution :

Tangent to the parabola $y={{x}^{2}}$at $(2,4)$is                   $\frac{1}{2}(y+4)=x.2$or $4x-y-4=0$                   It is also a tangent to the circle so that the centre lies on the normal through (2, 4) whose equation is $x+4y=\lambda ,$where $2+16=\lambda$                   $\therefore x+4y=18$is the normal on which lies $(h,k)$.                   \$h+4k=18$                                                 .....(i)                   Again distance of centre $(h,k)$from $(2,4)$and $(0,1)$on the circle are equal.                   $\therefore {{(h-2)}^{2}}+{{(k-4)}^{2}}={{h}^{2}}+{{(k-1)}^{2}}$                   $\therefore 4h+6k=19$                                                            .....(ii)                    Solving (i) and (ii), we get the centre$=\left( \frac{-16}{5},\frac{53}{10} \right)$.

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