If the circle `(x-6)^(2)+y^(2)=r^(2)` and the parabola `y^(2)=4x` have maximum number of common chords, then the least integral value of r is ________

Question

If the circle `(x-6)^(2)+y^(2)=r^(2)` and the parabola `y^(2)=4x` have maximum number of common chords, then the least integral value of r is ________

If the circle `(x-6)^(2)+y^(2)=r^(2)` and the parabola `y^(2)=4x` have maximum number of common chords, then the least integral value of r is __________ .

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 5
(5) For maximum number of common chords, the circle and the parabola must intersect at four distinct points.
Let us first find the of r when circle and the parabola touch each other.
For that, solving the given curves, we have
`(x-6)^(2)+4x=r^(2)`
`orx^(2)-8x+36-r^(2)=0`
The curves touch if discriminant
`D=64-4(36-r^(2))=0`
`orr^(2)=20`
Hence, the least integral value of r for which the curves intersect is 5.