Q. The centre of the circle passing through the point (0, 1)and touching the curve $y = x^2 at (2,4)$ is

Solution:

Let centre of circle be (h,k).
so that $\hspace10mm OA^2 = OB^2$
$\Rightarrow \hspace5mm h^2 + (R - 1)^2 = (h - 2)^2 + (R - 4)^2$
$\Rightarrow \hspace5mm 4h + 6R - 19 = 0$
Also, slope of OA $ = \frac {R - 4}{h - 2}$ 2 and slope of tangent at (2,4) to $y = x^2$ is 4.
and (slope of OA) . (slope of tangent at A) = - 1
$\therefore \hspace10mm \frac{R-4}{h-2} . 4 = - 1$
$\Rightarrow \hspace10mm 4R - 16 = - h + 2$
$\hspace15mm h + 4R = 18 \, \, \, \, \, \, \, \, ...(ii)$
On solving Eqs. (i) and (ii), we get
$\hspace20mm R = \frac {53}{10}$
and $\hspace16mm h = - \frac {16}{5}$
$\therefore Centre\, coordinates\, are \bigg( - \frac{16}{5}, \frac {53}{10} \bigg).$

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