Let a circle C in complex plane pass through the points z _ 1 = 3 + 4i, z _ 2 = 4 + 3i and z _ 3 = 5i . If z(≠ z _ 1 ) is a point on C such that the line through z and z _ 1 is perpendicular to the line through z _ 2

Correct answer: A. ^ - 1 ((24)/(7)) - π z _ 1 ≡ A ≡ (3, 4), z _ 2 ≡ B ≡ (4, 3), z _ 3 ≡ C ≡ (0, 5) and z ≡ P ≡ (x, y) Slope of BC = (3 - 5)/(4 - 0) = - (1)/(2) Slope of AP = 2 Now, equation of APisy - 4 = 2(x - 3) ⇒ y = 2(x - 1) Since P lies on circle x ^ 2 + y ^ 2 = 25 ⇒ x ^ 2 + (2(x - 1)) ^ 2 = 25 ⇒ x = - (7)/(5) and y = - (24)/(5) ⇒ arg ( z ) = ^ - 1 ((24)/(7)) - π

JEE Main 202225 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Complex Number

Let a circle C in complex plane pass through the points z _ 1 = 3 +...

Let a circle

C

in complex plane pass through the points

{z}_{1} = 3 + 4i, {z}_{2} = 4 + 3i

and

{z}_{3} = 5i

. If

z\left(\ne {z}_{1}\right)

is a point on

C

such that the line through

z

and

{z}_{1}

is perpendicular to the line through

{z}_{2}

and

{z}_{3}

, then

arg\left(z\right)

is equal to Let a circle C in complex plane pass through the points z _ 1 = 3 + 4i, z _ 2 = 4 + 3i and z _ 3 = 5i . If z(≠ z _ 1 )

Options

A $\tan^{- 1}\left(\dfrac{24}{7}\right) - \pi$
B $\tan^{- 1}\left(\dfrac{2}{\sqrt{5}}\right) - \pi$
C $\tan^{- 1}\left(3\right) - \pi$
D $\tan^{- 1}\left(\dfrac{3}{4}\right) - \pi$

Correct answer

A. $\tan^{- 1}\left(\dfrac{24}{7}\right) - \pi$

Step-by-step solution

${z}_{1} ≡ A ≡ \left(3, 4\right), {z}_{2} ≡ B ≡ \left(4, 3\right), {z}_{3} ≡ C ≡ \left(0, 5\right)$

and $z ≡ P ≡ \left(x, y\right)$

Let a circle C in complex plane pass through the points z _ 1 = 3 + 4i, z _ 2 = 4 + 3i and z _ 3 = 5i . If z(≠ z _ 1 )

Slope of $BC = \dfrac{3 - 5}{4 - 0} = - \dfrac{1}{2}$

Slope of $AP = 2$

Now, equation of $AP\text{is}y - 4 = 2\left(x - 3\right)$

$⇒ y = 2\left(x - 1\right)$

Since $P$ lies on circle ${x}^{2} + {y}^{2} = 25$

$⇒ {x}^{2} + {\left(2\left(x - 1\right)\right)}^{2} = 25$

$⇒ x = - \dfrac{7}{5}$ and $y = - \dfrac{24}{5}$

$⇒ arg ( z ) = \tan^{- 1}\left(\dfrac{24}{7}\right) - \pi$

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