The points of intersection of the line $\mathrm{ax}+$ by $=0$, $(\mathrm{a} \neq \mathrm{b})$ and the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}=0$ are $A(\alpha, 0)$ and $B(1, \beta)$. The image of the circle with AB as a diameter in the line $\mathrm{x}+\mathrm{y}+2=0$ is :

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Circle

The points of intersection of the line

ax +

= 0

(a \neq = b)

and the circle

x^{2} + y^{2} - 2 x = 0

are

A (α, 0)

and

B (1, β)

. The image of the circle with AB as a diameter in the line

x + y + 2 = 0

is :

Answer: (A)

The circle

x^{2} + y^{2} - 2 x = 0

passes through the origin, which also lies on

a x + b y = 0

, so one intersection must be the origin, forcing

A = (0, 0)

. Putting

x = 1

in the circle gives

β^{2} = 1

; the choice

β = 1

keeps

a \neq = b

, so

B = (1, 1)

. The circle on

A B

as diameter is

x^{2} + y^{2} - x - y = 0

with centre

(\frac{1}{2}, \frac{1}{2})

and radius

\frac{1}{2}

. Reflecting the centre in

x + y + 2 = 0

gives

(- \frac{5}{2}, - \frac{5}{2})

with the same radius, so the image circle is

x^{2} + y^{2} + 5 x + 5 y + 12 = 0

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option