JEE Advanced 1984

LEVELJEE Main

Animated Solution for Mathematics - Circles: The abscissa of the two points $A$ and $B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and their ordinates are the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Find the equation and the radius of the circle with $A B$ as diameter.

Q: Wait, why can't we just solve the quadratic equations to find the coordinates of $A$ and $B$ directly?

That is a great question! While you *could* solve for $x_1, x_2, y_1,$ and $y_2$ using the quadratic formula, it would lead to messy radical expressions. In JEE problems, we look for the most elegant path. By using Vieta's formulas, we can work directly with the sums and products of the roots, which simplifies the algebra significantly and reduces the chance of calculation errors.

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the circle on the coordinate plane. You are given the endpoints of the diameter,

A (x_{1}, y_{1})

and

B (x_{2}, y_{2})

. The key is to avoid finding the individual coordinates of

A

and

B

, which would be messy. Instead, focus on the relationship between the roots of the given quadratic equations and the coefficients of the circle's equation.

Hint 2

Hint 2: The Arsenal. Recall Vieta's formulas for a quadratic equation

x^{2} + B x + C = 0

. The sum of the roots is

- B

and the product of the roots is

C

. Apply this to both the

x

-coordinate equation (

x^{2} + 2 a x - b^{2} = 0

) and the

y

-coordinate equation (

y^{2} + 2 p y - q^{2} = 0

Hint 3

Hint 3: The Execution. Use the diameter form of the circle equation:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

. Expand this expression to get

x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} + y^{2} - (y_{1} + y_{2}) y + y_{1} y_{2} = 0

. Now, substitute the sums and products of the roots you found in the previous step.

Visualized Solution

Visualizing the Diameter $A B$

Let the endpoints of the diameter be $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ .
We need to find the equation of the circle with $A B$ as its diameter.

Roots for the $x$ -coordinates

The $x$ -coordinates $x_{1}$ and $x_{2}$ are the roots of the quadratic equation:
$x^{2} + 2 a x - b^{2} = 0$

Sum and Product of $x$ -coordinates

Using the properties of quadratic roots for $x^{2} + 2 a x - b^{2} = 0$ :
Sum of roots: $x_{1} + x_{2} = - 2 a$
Product of roots: $x_{1} x_{2} = - b^{2}$

Roots for the $y$ -coordinates

The $y$ -coordinates $y_{1}$ and $y_{2}$ are the roots of the quadratic equation:
$y^{2} + 2 p y - q^{2} = 0$

Sum and Product of $y$ -coordinates

Using the properties of quadratic roots for $y^{2} + 2 p y - q^{2} = 0$ :
Sum of roots: $y_{1} + y_{2} = - 2 p$
Product of roots: $y_{1} y_{2} = - q^{2}$

The Diameter Form of a Circle

The equation of a circle with diameter endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:
$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

Expanding the Diameter Form

Expanding the brackets:
$x^{2} - x_{1} x - x_{2} x + x_{1} x_{2} + y^{2} - y_{1} y - y_{2} y + y_{1} y_{2} = 0$
Grouping the terms:
$x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} + y^{2} - (y_{1} + y_{2}) y + y_{1} y_{2} = 0$

Substituting Sums and Products

Substituting $x_{1} + x_{2} = - 2 a$ , $x_{1} x_{2} = - b^{2}$ , $y_{1} + y_{2} = - 2 p$ , and $y_{1} y_{2} = - q^{2}$ :
$x^{2} - (- 2 a) x + (- b^{2}) + y^{2} - (- 2 p) y + (- q^{2}) = 0$

Simplifying the Equation

Simplifying the signs:
$x^{2} + 2 a x - b^{2} + y^{2} + 2 p y - q^{2} = 0$
Rearranging into standard form:
$x^{2} + y^{2} + 2 a x + 2 p y - b^{2} - q^{2} = 0$

Comparing with General Form

The general form of a circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Comparing this with our equation:
$g = a$
$f = p$
$c = - (b^{2} + q^{2})$

Applying the Radius Formula

The radius $r$ is given by the formula:
$r = g^{2} + f^{2} - c$
Substituting our values:
$r = a^{2} + p^{2} - (- (b^{2} + q^{2}))$
$r = a^{2} + p^{2} + b^{2} + q^{2}$

Final Answer

Equation of the circle: $x^{2} + y^{2} + 2 a x + 2 p y - b^{2} - q^{2} = 0$
Radius of the circle: $a^{2} + p^{2} + b^{2} + q^{2}$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

The problem defines a circle where the diameter is determined by two points,

A

and

B

. The abscissae (

x

-coordinates) of these points are the roots of the quadratic equation:

x^{2} + 2 a x - b^{2} = 0

Similarly, the ordinates (

y

-coordinates) of these points are the roots of the quadratic equation:

y^{2} + 2 p y - q^{2} = 0

The Hidden Coordinates

We define the roots of the

x

-equation as

x_{1}

and

x_{2}

, and the roots of the

y

-equation as

y_{1}

and

y_{2}

. Rather than solving for these values individually, we utilize Vieta's formulas to extract the necessary information.

For the

x

-coordinates:

x_{1} + x_{2} = - 2 a

x_{1} x_{2} = - b^{2}

For the

y

-coordinates:

y_{1} + y_{2} = - 2 p

y_{1} y_{2} = - q^{2}

The Diameter Form

When the endpoints of a diameter are known as

(x_{1}, y_{1})

and

(x_{2}, y_{2})

, the circle is defined by the diameter form equation:

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

Expanding this expression yields:

x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} + y^{2} - (y_{1} + y_{2}) y + y_{1} y_{2} = 0

The Synthesis

By substituting the values derived from Vieta's formulas into the expanded equation, we obtain:

x^{2} - (- 2 a) x + (- b^{2}) + y^{2} - (- 2 p) y + (- q^{2}) = 0

Simplifying the signs, we arrive at the final equation of the circle:

x^{2} + y^{2} + 2 a x + 2 p y - b^{2} - q^{2} = 0

Final Calculation of the Radius

To find the radius, we compare our result to the general form of a circle,

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

. By inspection, we identify the parameters:

g = a, f = p, c = - (b^{2} + q^{2})

The radius

r

is calculated using the standard formula

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

. Substituting our identified parameters:

r = a^{2} + p^{2} - (- (b^{2} + q^{2}))

Thus, the final radius is:

r = a^{2} + p^{2} + b^{2} + q^{2}

Common Traps & FAQs

Wait, why can't we just solve the quadratic equations to find the coordinates of

A

and

B

directly?

That is a great question! While you *could* solve for

x_{1}, x_{2}, y_{1},

and

y_{2}

using the quadratic formula, it would lead to messy radical expressions. In JEE problems, we look for the most elegant path. By using Vieta's formulas, we can work directly with the sums and products of the roots, which simplifies the algebra significantly and reduces the chance of calculation errors.

The problem gives the

y

-coordinate equation as

x^{2} + 2 p x - q^{2} = 0

. Is that a typo? Should it be

y^{2}

It is not a typo! In algebra, the variable used in a quadratic equation is just a placeholder or a 'dummy variable.' Whether the equation is written in terms of

x

y

, the roots represent the values of the ordinates. You can safely treat it as

y^{2} + 2 p y - q^{2} = 0

to avoid confusion.

How do I know when to use the diameter form of the circle equation versus the center-radius form?

The choice depends entirely on the information given. If you are given the center

(h, k)

and the radius

r

, the center-radius form

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

is best. However, when you are given the endpoints of a diameter, the diameter form

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0

is the most direct and powerful tool in your arsenal.

Animated Solution for Mathematics - Circles: The abscissa of the two points $A$ and $B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and their ordinates are the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Find the equation and the radius of the circle with $A B$ as diameter.

Visualized Solution

Visualizing the Diameter $A B$

Roots for the $x$ -coordinates

Sum and Product of $x$ -coordinates

Roots for the $y$ -coordinates

Sum and Product of $y$ -coordinates

The Diameter Form of a Circle

Expanding the Diameter Form

Substituting Sums and Products

Simplifying the Equation

Comparing with General Form

Applying the Radius Formula

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Coordinates

The Diameter Form

The Synthesis

Final Calculation of the Radius

Similar Questions

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Visualized Solution

Visualizing the Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Roots for the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinates

Sum and Product of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinates

Roots for the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates

Sum and Product of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates

The Diameter Form of a Circle

Expanding the Diameter Form

Substituting Sums and Products

Simplifying the Equation

Comparing with General Form

Applying the Radius Formula

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Coordinates

The Diameter Form

The Synthesis

Final Calculation of the Radius

Similar Questions

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Sum and Product of $x$ -coordinates

Roots for the $y$ -coordinates

Sum and Product of $y$ -coordinates

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