Why must instead of ?

If , the line passing through forces . Then for to lie on , , which requires (since ). But is invalid for a line, and violates .

JEE MainMathematics

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Circles: JEE Main Mathematics Question with Solution

The points of intersection of the line

ax+by=0,(a \neq b)

and the circle

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

are

\mathrm{A}(\alpha, 0)

and

\mathrm{B}(1, \beta)

. The image of the circle with

\mathrm{AB}

as a diameter in the line

x+y+2=0

is:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

x^2+y^2+3x+3y+4=0

x^2+y^2+3x+5y+8=0

x^2+y^2-5x-5y+12=0

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

Step-by-step solutionView

Correct answer

The intersection points are $A(0,0)$ and $B(1,1)$ , giving a circle centered at $(1/2, 1/2)$ with radius $1/\sqrt{2}$ ; reflecting this center across $x+y+2=0$ yields $(-5/2, -5/2)$ , which corresponds to the circle $x^2+y^2+5x+5y+12=0$ .

Option analysis

Why each option works or fails

A ·

x^2+y^2+3x+3y+4=0

Believing that the reflected center has coordinates $(-3/2, -3/2)$ by omitting the constant term or misapplying the reflection formula. Use the standard reflection formula $\frac{x'-x_0}{a} = \frac{y'-y_0}{b} = \frac{-2(ax_0+by_0+c)}{a^2+b^2}$ carefully with $a=1, b=1, c=2$ .

B ·

x^2+y^2+3x+5y+8=0

Assuming the reflected circle does not retain symmetry along the line $y=x$ , leading to unequal linear coefficients in $x$ and $y$ . Notice that because the original center $(1/2, 1/2)$ lies on the line of symmetry $y=x$ perpendicular to $x+y+2=0$ , the reflected center must also satisfy $x'=y'$ .

C ·

x^2+y^2-5x-5y+12=0

Inverting the sign of the reflected coordinates, yielding center $(5/2, 5/2)$ instead of $(-5/2, -5/2)$ . Check the geometry: the point $(1/2, 1/2)$ reflected across $x+y=-2$ must move into the third quadrant where both coordinates are negative.

D ·

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

None. This is the correct equation of the reflected circle. This equation correctly combines the reflected center $(-5/2, -5/2)$ and the unchanged radius $r = 1/\sqrt{2}$ .

Reviewed route

Solution

StepWorking

01given

Line $ax+by=0$ with $a \neq b$ , circle $C_1: x^2+y^2-2x=0$ . Intersection points are $A(\alpha, 0)$ and $B(1, \beta)$ . Line of reflection is $L: x+y+2=0$ .

02goal

Find the equation of the image of the circle having diameter $AB$ in the line $x+y+2=0$ .

03approach

Substitute the coordinates of $A$ and $B$ into the circle equation and the line equation to uniquely determine $\alpha$ and $\beta$ . Then form the circle with diameter $AB$ , find its center and radius, compute the image of the center in the line $x+y+2=0$ , and construct the reflected circle.

04execute

Point $A(\alpha, 0)$ lies on circle $x^2+y^2-2x=0 \implies \alpha^2 - 2\alpha = 0 \implies \alpha = 0$ or $\alpha = 2$ . Since $A$ also lies on $ax+by=0$ , we have $a\alpha = 0$ . If $\alpha = 2$ , then $a = 0$ . Point $B(1, \beta)$ lies on $x^2+y^2-2x=0 \implies 1 + \beta^2 - 2(1) = 0 \implies \beta^2 = 1 \implies \beta = \pm 1$ . Also, $B$ is on $ax+by=0 \implies a(1) + b\beta = 0$ . If $a=0$ , then $b\beta=0$ , which would mean $b=0$ since $\beta = \pm 1$ , contradicting $a \neq b$ . Thus $\alpha = 0$ , giving $A(0,0)$ . Then $a(1)+b\beta=0 \implies \beta = -a/b$ . From $\beta = \pm 1$ , if $\beta = -1$ , $a = b$ , which contradicts $a \neq b$ . Therefore, $\beta = 1$ (and $a = -b$ ). Thus, $A=(0,0)$ and $B=(1,1)$ .

05execute

The circle with diameter $AB$ has equation $(x-0)(x-1) + (y-0)(y-1) = 0 \implies x^2+y^2-x-y=0$ . Its center is $C = (1/2, 1/2)$ and radius squared is $r^2 = (1/2)^2 + (1/2)^2 = 1/2$ .

06execute

Let the image of center $(h_0, k_0) = (1/2, 1/2)$ in the line $x+y+2=0$ be $(h', k')$ . Using the reflection formula $\frac{h'-1/2}{1} = \frac{k'-1/2}{1} = -2\frac{1/2 + 1/2 + 2}{1^2+1^2} = -2\frac{3}{2} = -3$ . Thus $h' = 1/2 - 3 = -5/2$ and $k' = 1/2 - 3 = -5/2$ .

07execute

The reflected circle preserves radius, so its equation is $(x + 5/2)^2 + (y + 5/2)^2 = 1/2 \implies x^2 + 5x + 25/4 + y^2 + 5y + 25/4 = 1/2 \implies x^2+y^2+5x+5y + 25/2 - 1/2 = 0 \implies x^2+y^2+5x+5y+12=0$ .

✓verify

Check that the radius is unchanged: $g = 5/2, f = 5/2, c = 12 \implies r = \sqrt{(5/2)^2 + (5/2)^2 - 12} = \sqrt{25/2 - 12} = \sqrt{1/2}$ , which matches the original radius $\sqrt{1/2}$ .

Hints that build this answer step by step

What are the coordinates of the intersection points $A(\alpha, 0)$ and $B(1, \beta)$ ?
$A(0,0)$ and $B(1,1)$
What are the center $C$ and radius $r$ of the circle with diameter $AB$ , where $A=(0,0)$ and $B=(1,1)$ ?
Center $C = \left(\frac{1}{2}, \frac{1}{2}\right)$ , radius $r = \frac{1}{\sqrt{2}}$
What is the reflection of the center $C\left(\frac{1}{2}, \frac{1}{2}\right)$ across the line $x+y+2=0$ ?
$C' = \left(-\frac{5}{2}, -\frac{5}{2}\right)$
What is the equation of the reflected circle with center $\left(-\frac{5}{2}, -\frac{5}{2}\right)$ and radius $r = \frac{1}{\sqrt{2}}$ ?
$x^2+y^2+5x+5y+12=0$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why must

\alpha = 0

instead of

2

If $\alpha = 2$ , the line $ax+by=0$ passing through $(2,0)$ forces $a=0$ . Then for $(1,\beta)$ to lie on $ax+by=0$ , $b\beta=0$ , which requires $b=0$ (since $\beta = \pm 1 \neq 0$ ). But $a=b=0$ is invalid for a line, and violates $a \neq b$ .

Answer

The intersection points are $A(0,0)$ and $B(1,1)$ , giving a circle centered at $(1/2, 1/2)$ with radius $1/\sqrt{2}$ ; reflecting this center across $x+y+2=0$ yields $(-5/2, -5/2)$ , which corresponds to the circle $x^2+y^2+5x+5y+12=0$ .

Why each option works or fails

A: $x^2+y^2+3x+3y+4=0$ - Believing that the reflected center has coordinates $(-3/2, -3/2)$ by omitting the constant term or misapplying the reflection formula. Use the standard reflection formula $\frac{x'-x_0}{a} = \frac{y'-y_0}{b} = \frac{-2(ax_0+by_0+c)}{a^2+b^2}$ carefully with $a=1, b=1, c=2$ .
B: $x^2+y^2+3x+5y+8=0$ - Assuming the reflected circle does not retain symmetry along the line $y=x$ , leading to unequal linear coefficients in $x$ and $y$ . Notice that because the original center $(1/2, 1/2)$ lies on the line of symmetry $y=x$ perpendicular to $x+y+2=0$ , the reflected center must also satisfy $x'=y'$ .
C: $x^2+y^2-5x-5y+12=0$ - Inverting the sign of the reflected coordinates, yielding center $(5/2, 5/2)$ instead of $(-5/2, -5/2)$ . Check the geometry: the point $(1/2, 1/2)$ reflected across $x+y=-2$ must move into the third quadrant where both coordinates are negative.
D · correct: $x^2+y^2+5x+5y+12=0$ - None. This is the correct equation of the reflected circle. This equation correctly combines the reflected center $(-5/2, -5/2)$ and the unchanged radius $r = 1/\sqrt{2}$ .

Step-by-step solution

given: Line $ax+by=0$ with $a \neq b$ , circle $C_1: x^2+y^2-2x=0$ . Intersection points are $A(\alpha, 0)$ and $B(1, \beta)$ . Line of reflection is $L: x+y+2=0$ .
goal: Find the equation of the image of the circle having diameter $AB$ in the line $x+y+2=0$ .
approach: Substitute the coordinates of $A$ and $B$ into the circle equation and the line equation to uniquely determine $\alpha$ and $\beta$ . Then form the circle with diameter $AB$ , find its center and radius, compute the image of the center in the line $x+y+2=0$ , and construct the reflected circle.
execute: Point $A(\alpha, 0)$ lies on circle $x^2+y^2-2x=0 \implies \alpha^2 - 2\alpha = 0 \implies \alpha = 0$ or $\alpha = 2$ . Since $A$ also lies on $ax+by=0$ , we have $a\alpha = 0$ . If $\alpha = 2$ , then $a = 0$ . Point $B(1, \beta)$ lies on $x^2+y^2-2x=0 \implies 1 + \beta^2 - 2(1) = 0 \implies \beta^2 = 1 \implies \beta = \pm 1$ . Also, $B$ is on $ax+by=0 \implies a(1) + b\beta = 0$ . If $a=0$ , then $b\beta=0$ , which would mean $b=0$ since $\beta = \pm 1$ , contradicting $a \neq b$ . Thus $\alpha = 0$ , giving $A(0,0)$ . Then $a(1)+b\beta=0 \implies \beta = -a/b$ . From $\beta = \pm 1$ , if $\beta = -1$ , $a = b$ , which contradicts $a \neq b$ . Therefore, $\beta = 1$ (and $A(\alpha, 0)$ ). Thus, $A=(0,0)$ and $B=(1,1)$ .
execute: The circle with diameter $AB$ has equation $(x-0)(x-1) + (y-0)(y-1) = 0 \implies x^2+y^2-x-y=0$ . Its center is $C = (1/2, 1/2)$ and radius squared is $x^2+y^2-2x=0 \implies 1 + \beta^2 - 2(1) = 0 \implies \beta^2 = 1 \implies \beta = \pm 1$ .
execute: Let the image of center $(h_0, k_0) = (1/2, 1/2)$ in the line $x+y+2=0$ be $(h', k')$ . Using the reflection formula $\frac{h'-1/2}{1} = \frac{k'-1/2}{1} = -2\frac{1/2 + 1/2 + 2}{1^2+1^2} = -2\frac{3}{2} = -3$ . Thus $h' = 1/2 - 3 = -5/2$ and $k' = 1/2 - 3 = -5/2$ .
execute: The reflected circle preserves radius, so its equation is $(x + 5/2)^2 + (y + 5/2)^2 = 1/2 \implies x^2 + 5x + 25/4 + y^2 + 5y + 25/4 = 1/2 \implies x^2+y^2+5x+5y + 25/2 - 1/2 = 0 \implies x^2+y^2+5x+5y+12=0$ .
verify: Check that the radius is unchanged: $g = 5/2, f = 5/2, c = 12 \implies r = \sqrt{(5/2)^2 + (5/2)^2 - 12} = \sqrt{25/2 - 12} = \sqrt{1/2}$ , which matches the original radius $\sqrt{1/2}$ .

Circles: JEE Main Mathematics Question with Solution

What feels right?

The intersection points are $A(0,0)$ and $B(1,1)$ , giving a circle centered at $(1/2, 1/2)$ with radius $1/\sqrt{2}$ ; reflecting this center across $x+y+2=0$ yields $(-5/2, -5/2)$ , which corresponds to the circle $x^2+y^2+5x+5y+12=0$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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