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

Options

A. `(-(16)/(5),(27)/(10))`
B. `(-(16)/(7),(53)/(10))`
C. `(-(16)/(5),(53)/(10))`
D. None of the above

Answer

C

Explanation

Let centre of circle be (h,k). So that `OA^(2)=OB^(2)`

`rArr h^(2)+(k-1)^(2)=(h-2)^(2)+(k-4)^(2)`
`rArr 4h +6k - 19 = 0 ` ...(i)
Also, slope of OA = `(k-4)/(h-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 (k-4)/(h-2).4=-1`
`rArr 4k - 16 = -h+2`
h + 4k = 18 ...(ii)
On solving Eqs. (i) and (ii), we get
`k = (53)/(10) and h = -(16)/(5)`
`therefore` Centre coordinates are `(-(16)/(5), (53)/(10))`.