• # question_answer 13) The centre of the circle passing through the point $\left( \mathbf{0},\mathbf{1} \right)$ and touching the curve $\mathbf{y}={{\mathbf{x}}^{\mathbf{2}}}$ at $\left( \mathbf{2},\mathbf{4} \right)$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) $\left( \frac{16}{5},\frac{-53}{10} \right)$

[c] Let the centre of circle be $\left( \alpha ,\beta \right)$ $\therefore O{{A}^{2}}=O{{B}^{2}}$ ${{\alpha }^{2}}+{{(\beta -1)}^{2}}={{(\beta -2)}^{2}}+{{(\beta -4)}^{2}}$ ${{\alpha }^{2}}+{{\beta }^{2}}-2\beta +1$ $={{\alpha }^{2}}-4\alpha +4+{{\beta }^{2}}-8\beta +16$ $=4\alpha +6\beta -19=0$                       .......(1) Slope of $OA=\frac{\beta -4}{\alpha -2}$ Slope of tangent at $\left( 2,4 \right)$ to $y={{x}^{2}}$be ${{y}_{1}}at(2,4)=2x=2\times 2=4$ Since, (slope of OA) (slope of tangent at $(A)=-1$ $\frac{\beta -1}{\alpha -2}\times 4=-1$ $\Rightarrow 4\beta -16=-\alpha +2$ $\Rightarrow \alpha +4\beta -18$                        ??...,.(2) Solving equation (1) and (2), we have $\frac{\alpha }{108-76}=\frac{\beta }{19-72}=\frac{-1}{16-6}$ $\therefore \alpha =\frac{-32}{10}=\frac{16}{5}$ $\beta =\frac{53}{10}$Thus, centre of the circle, $\left( \alpha ,\beta \right)=\left( \frac{-16}{5},\frac{53}{10} \right)$ Hence, option [c]is correct