I am confused about the relationship between $\gamma$ and $a$. Why is $\gamma = a^{2}$?

Great question. When we expanded the equation $(x-a)^{2} + (y+r)^{2} = r^{2}$, we obtained $x^{2} + y^{2} - 2ax + 2ry + a^{2} = 0$. Comparing this to the general form $x^{2} + y^{2} - \alpha x + \beta y + \gamma = 0$, the constant term $\gamma$ must correspond to $a^{2}$. It is a direct consequence of the circle touching the x-axis at $x=a$.

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let the equation of the circle, which touches x-axis at the point $(a, 0)$ , $a > 0$ and cuts off an intercept of length b on y-axis be $x^{2} + y^{2} - α x + β y + γ = 0$ If the circle lies below x-axis, then the ordered pair $(2 a, b^{2})$ is equal to

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry first. If a circle touches the x-axis at

(a, 0)

and resides entirely below the axis, its center must be located at

(a, - r)

, where

r

is the radius. Why? Because the distance from the center to the point of tangency must be the radius, and the vertical drop is necessary to stay below the axis.

Hint 2

Now, bring in your algebraic toolkit. The general equation of a circle is

x^{2} + y^{2} - α x + β y + γ = 0

. Recall that for any circle

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, the center is

(- g, - f)

and the radius is

g^{2} + f^{2} - c

. Use this to map your specific parameters

α, β, γ

to the geometric center and radius.

Hint 3

Finally, tackle the y-intercept. The length of the intercept cut by a circle on the y-axis is given by

2 f^{2} - c

. In your general equation, identify

f

in terms of

β

and

c

in terms of

γ

. Square the intercept length

b

to eliminate the radical and find the direct relationship between

b^{2}, β,

and

γ

Visualized Solution

Visualizing the Circle Position

Circle touches the x-axis at $(a, 0)$ where $a > 0$ .
The circle lies below the x-axis.
Therefore, the center must be $(a, - r)$ and the radius is $r$ .

Standard Equation of a Circle

Standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$
We know the center $(h, k) = (a, - r)$
The radius $R = r$

Substituting Center and Radius

Substitute $h = a$ , $k = - r$ , and $R = r$ :
$(x - a)^{2} + (y - (- r))^{2} = r^{2}$
$(x - a)^{2} + (y + r)^{2} = r^{2}$

Expanding the Equation

Expand $(x - a)^{2}$ : $x^{2} - 2 a x + a^{2}$
Expand $(y + r)^{2}$ : $y^{2} + 2 r y + r^{2}$
Combine: $x^{2} - 2 a x + a^{2} + y^{2} + 2 r y + r^{2} = r^{2}$

Simplifying the Equation

Cancel $r^{2}$ from both sides.
Rearrange terms: $x^{2} + y^{2} - 2 a x + 2 r y + a^{2} = 0$

Comparing with Given Equation

Given equation: $x^{2} + y^{2} - α x + β y + γ = 0$
Our equation: $x^{2} + y^{2} - 2 a x + 2 r y + a^{2} = 0$
Comparing coefficients:
$α = 2 a$
$β = 2 r$
$γ = a^{2}$

Analyzing the Y-intercept

The circle cuts off an intercept of length $b$ on the y-axis.
Formula for y-intercept length: $2 f^{2} - c$
In our general equation, $f = \frac{β}{2}$ and $c = γ$ .

Substituting into Intercept Formula

From our comparison: $f = \frac{β}{2} = r$
And the constant $c = γ = a^{2}$
Substitute these into the intercept formula:
$b = 2 r^{2} - a^{2}$

Squaring the Intercept Equation

We have $b = 2 r^{2} - a^{2}$
Square both sides to remove the square root:
$b^{2} = 4 (r^{2} - a^{2})$
$b^{2} = 4 r^{2} - 4 a^{2}$

Expressing $b^{2}$ in terms of $β$ and $γ$

We know $β = 2 r ⟹ β^{2} = 4 r^{2}$
We know $γ = a^{2} ⟹ 4 γ = 4 a^{2}$
Substitute these into $b^{2} = 4 r^{2} - 4 a^{2}$ :
$b^{2} = β^{2} - 4 γ$

Final Ordered Pair

The question asks for the ordered pair $(2 a, b^{2})$ .
From Step 5, we know $2 a = α$ .
From Step 9, we know $b^{2} = β^{2} - 4 γ$ .
Therefore, $(2 a, b^{2}) = (α, β^{2} - 4 γ)$ .

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Geometric Setup

We are given a circle that touches the

x

-axis at the point

(a, 0)

, where

a > 0

. Since the circle lies entirely below the

x

-axis, its center must be located directly below the point of tangency.

Given the radius

r

, the center of the circle is at

(a, - r)

. This configuration ensures the circle is tangent to the

x

-axis while remaining in the lower half-plane.

The Master Equation

Using the standard form of a circle equation,

(x - h)^{2} + (y - k)^{2} = R^{2}

, we substitute our center

(a, - r)

and radius

r

(x - a)^{2} + (y - (- r))^{2} = r^{2}

Expanding this expression, we obtain:

(x - a)^{2} + (y + r)^{2} = r^{2}

x^{2} - 2 a x + a^{2} + y^{2} + 2 r y + r^{2} = r^{2}

After canceling the

r^{2}

terms, the equation simplifies to:

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

Bridging to the General Form

We compare our derived equation to the general form provided:

x^{2} + y^{2} - α x + β y + γ = 0

. By matching the coefficients, we identify the following relationships:

α = 2 a

β = 2 r

γ = a^{2}

Calculating the Y-Intercept

To find the intercept of length

b

on the

y

-axis, we set

x = 0

in the general equation:

y^{2} + β y + γ = 0

The roots of this quadratic equation,

y_{1}

and

y_{2}

, represent the

y

-coordinates of the intersection points. The length of the intercept is given by

∣ y_{1} - y_{2} ∣ = b

Using the properties of quadratic roots, we know:

b = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2} = β^{2} - 4 γ

Squaring both sides, we find:

b^{2} = β^{2} - 4 γ

Final Result

The problem asks for the ordered pair

(2 a, b^{2})

. Substituting our derived values:

2 a = α

b^{2} = β^{2} - 4 γ

Thus, the final result is:

$(α, β^{2} - 4 γ)$

Common Traps & FAQs

Wait, why is the center

(a, - r)

and not

(a, r)

? If the circle touches the x-axis, shouldn't the y-coordinate be positive?

That is a fantastic observation! If the circle were above the x-axis, the center would indeed be

(a, r)

. However, the problem explicitly states the circle lies *below* the x-axis. To touch the x-axis at

(a, 0)

while remaining below it, the circle must 'hang' downwards. Thus, the center must be at a negative y-coordinate, specifically

- r

Why is the y-intercept formula

2 f^{2} - c

? Can't I just set

x = 0

in the circle equation?

You absolutely can, and that is the most robust way to derive it! If you set

x = 0

x^{2} + y^{2} - α x + β y + γ = 0

, you get a quadratic in

y

y^{2} + β y + γ = 0

. The roots

y_{1}

and

y_{2}

are the points where the circle cuts the y-axis. The length of the intercept is

∣ y_{1} - y_{2} ∣

. Using the property

∣ y_{1} - y_{2} ∣ = (y_{1} + y_{2})^{2} - 4 y_{1} y_{2}

, you will arrive at the same formula. It is beautiful how the algebra confirms the geometry!

I am confused about the relationship between

γ

and

a

. Why is

γ = a^{2}

Great question. When we expanded the equation

(x - a)^{2} + (y + r)^{2} = r^{2}

, we obtained

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

. Comparing this to the general form

x^{2} + y^{2} - α x + β y + γ = 0

, the constant term

γ

must correspond to

a^{2}

. It is a direct consequence of the circle touching the x-axis at

x = a

Select Answer:

Visualized Solution

Visualizing the Circle Position

Standard Equation of a Circle

Substituting Center and Radius

Expanding the Equation

Simplifying the Equation

Comparing with Given Equation

Analyzing the Y-intercept

Substituting into Intercept Formula

Squaring the Intercept Equation

Expressing $b^{2}$ in terms of $β$ and $γ$

Final Ordered Pair

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Geometric Setup

The Master Equation

Bridging to the General Form

Calculating the Y-Intercept

Final Result

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Select Answer:

Visualized Solution

Visualizing the Circle Position

Standard Equation of a Circle

Substituting Center and Radius

Expanding the Equation

Simplifying the Equation

Comparing with Given Equation

Analyzing the Y-intercept

Substituting into Intercept Formula

Squaring the Intercept Equation

Final Ordered Pair

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Geometric Setup

The Master Equation

Bridging to the General Form

Calculating the Y-Intercept

Final Result

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