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JEE 2023 Mathematics PYQ — A circle passing through the point P(a, b) in the first quadrant … | Mathem Solvex | Mathem Solvex

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

A circle passing through the point P(a, b) in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then value of ab is ____.

Choose the correct answer:

A.
120
B.
121
(Correct Answer)
C.
122
D.
123

Correct Answer:

121

Explanation

Let the equation of the circle
(x – a)² + (y – a)² = a²

Which passes through P (α, β) then
$(α - a)^{2} + (β - a)^{2} = a^{2}$
$\Rightarrow α^{2} + β^{2} - 2 aa - 2 β a + a^{2} = 0$

Here equation of AB is $x + y = a$

Let Q (α, β) be foot of the perpendicular of p on AB

$\frac{α ^{'} - α}{1} = \frac{β ^{'} - β}{1} = \frac{- ( α + β - a )}{2}$

PQ² = $(α^{'} - α)^{2} + (β^{'} - β)^{2} = \frac{1}{4} (α + β - a)^{2} + \frac{1}{4} (α + β - a)^{2}$

$\Rightarrow 121 = \frac{1}{2} (α + β - a)^{2}$

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Explanation

Let the equation of the circle
(x – a)² + (y – a)² = a²

Which passes through P (α, β) then
$(α - a)^{2} + (β - a)^{2} = a^{2}$
$\Rightarrow α^{2} + β^{2} - 2 aa - 2 β a + a^{2} = 0$

Here equation of AB is $x + y = a$

Let Q (α, β) be foot of the perpendicular of p on AB

$\frac{α ^{'} - α}{1} = \frac{β ^{'} - β}{1} = \frac{- ( α + β - a )}{2}$

PQ² = $(α^{'} - α)^{2} + (β^{'} - β)^{2} = \frac{1}{4} (α + β - a)^{2} + \frac{1}{4} (α + β - a)^{2}$

$\Rightarrow 121 = \frac{1}{2} (α + β - a)^{2}$

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