JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Circle

Let PQ be a diameter of the circle

x^{2} + y^{2} = 9

. If

α

and

β

are the lengths of the perpendiculars from P and Q on the straight line,

x + y = 2

respectively, then the maximum value of

α β

Answer: 7

Solution

Let

P = (3 cos θ, 3 sin θ)

and the diametrically opposite point

Q = (- 3 cos θ, - 3 sin θ)

x^{2} + y^{2} = 9

. Their perpendicular distances to

x + y - 2 = 0

are

α = \frac{∣3 cos θ + 3 sin θ - 2∣}{2}

and

β = \frac{∣3 cos θ + 3 sin θ + 2∣}{2}

. Putting

s = 3 cos θ + 3 sin θ

α β = \frac{∣ s ^{2} - 4∣}{2}

, and since

s = 32 sin (θ + \frac{π}{4})

gives

s^{2} \in [0, 18]

, the maximum is

\frac{18 - 4}{2} = 7

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer

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