Parabola

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

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Solution

For maximum number of common chords, circle and parabola
must intersect in 4 distinct points.
Let's first find the value of r when circle and parabola touch each other.
For that solving the given curves we have
$\begin{aligned} \begin{aligned} (x - 6)^{2} + 4 x = r^{2} \\ or x^{2} - 8 x + 36 - r^{2} = 0 \end{aligned} \\ Curves touch if discriminant is 0 . \end{aligned}$
$D = 64 - 4 (36 - r^{2}) = 0 or r^{2} = 20$
Hence least integral value of r for which the curves intersect is 5.

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Parabola

Remember concepts with our Masterclasses.

If circle (x−6)2+y2=r2 and parabola y2=4x have maximum number of common chords, then least integral value of r is

Ready to Test Your Skills?

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