Let a circle C in complex plane pass through the points $\mathrm{z}_{1}=3+4 \mathrm{i}, \mathrm{z}_{2}=4+3 \mathrm{i}$ and $\mathrm{z}_{3}=5 \mathrm{i}$. If $\mathrm{z}\left(\neq \mathrm{z}_{1}\right)$ is a point on C such that the line through z and $\mathrm{z}_{1}$ is perpendicular to the line through $\mathrm{z}_{2}$ and $\mathrm{z}_{3}$, then $\arg (\mathrm{z})$ is equal to:

(B) $\tan ^{-1}\left(\frac{24}{7}\right)-\pi$

(A) $\tan ^{-1}\left(\frac{2}{\sqrt{5}}\right)-\pi$

(D) $\tan ^{-1}\left(\frac{3}{4}\right)-\pi$

Complex Numbers — JEE Main · 2022 · 25 Jun 2022 · Shift 1 Question with Solution

Let a circle C in complex plane pass through the points

z_{1} = 3 + 4 i, z_{2} = 4 + 3 i

and

z_{3} = 5 i

. If

z (\neq = z_{1})

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

ar g (z)

is equal to:

Answer: (B)

The three points

z_{1} = 3 + 4 i

z_{2} = 4 + 3 i

z_{3} = 5 i

correspond to

(3, 4)

(4, 3)

(0, 5)

in Cartesian coordinates.

The circle equation is

x^{2} + y^{2} = 25

(can be verified by substitution).

Slope of line through

z_{2}

and

z_{3}

m_{B C} = \frac{3 - 5}{4 - 0} = - \frac{1}{2}

Since line through

z

and

z_{1}

is perpendicular to this, its slope is

2

.

Line through

A = (3, 4)

with slope

2

y - 4 = 2 (x - 3)

, so

y = 2 x - 2

.

Substitute into circle equation:

x^{2} + (2 x - 2)^{2} = 25

x^{2} + 4 x^{2} - 8 x + 4 = 25

5 x^{2} - 8 x - 21 = 0

x = \frac{8 \pm 64 + 420}{10} = \frac{8 \pm 22}{10}

x = 3

(gives

z_{1}

) or

x = - \frac{7}{5}

For

x = - \frac{7}{5}

y = 2 (- \frac{7}{5}) - 2 = - \frac{24}{5}

z = - \frac{7}{5} - \frac{24}{5} i

(third quadrant)

ar g (z) = tan^{- 1} (\frac{24/5}{7/5}) - π = tan^{- 1} (\frac{24}{7}) - π

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option