what-is-the-centre-of-the-circle-with-radius-4-2-that-can-be-inscribed-in-the-parabola-y-x-2-16x-128-

Question Number 117739 by bemath last updated on 13/Oct/20

what is the centre of the circle

with radius 4 \sqrt{2} that can be

inscribed in the parabola

y = x^{2} - 16 x + 128 ?

Answered by bobhans last updated on 13/Oct/20

for symetry reasons, the center of circle

will lie on the axis of the parabola, say

it center is (8, u) and the equation is

{(x - 8)}^{2} + {(y - u)}^{2} = {(4 \sqrt{2})}^{2}

{(x - 8)}^{2} + {(y - u)}^{2} = 32

x^{2} - 16 x = - {(y - u)}^{2} - 32

if we subtitute in the equation of the

parabola gives y = - {(y - u)}^{2} - 32

or y^{2} + (1 - 2 u) y + u^{2} - 96 = 0

this should be have one roots, so we

get {(1 - 2 u)}^{2} - 4 (u^{2} - 96) = 0

1 - 4 u + {4 u}^{2} - {4 u}^{2} + 384 = 0

\Leftrightarrow u = \frac{385}{4} . Thus the center of the circle

is (8, \frac{385}{4})

Commented by bemath last updated on 13/Oct/20

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