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Question: The centre of the circle passing through the point (0, 1) and touching the curve y = x<sup>2</sup> a...

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

A

(-16/5, 27/10)

B

(-16/7, 5/10)

C

(-16/5, 53/10)

D

None of these

Answer

(-16/5, 53/10)

Explanation

Solution

Tangent to the parabola y = x2 at (2, 4) is

12(y+4)=x.26muor6mu4xy4=0\frac{1}{2}(y + 4) = x.2\mspace{6mu} or\mspace{6mu} 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 = λ, where

2 + 16 = λ.

∴ 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.

∴ (h-2)2 + (k-4)2 = h2 + (k –1)2

∴ 4h+6k = 19 ..(ii)

Solving (i) and (ii), we get the centre =(-16/5, 53/10)