JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: The equation of a circle is $Re (z^{2}) + 2 (Im (z))^{2} + 2 Re (z) = 0$ , where $z = x + i y$ . A line which passes through the center of the given circle and the vertex of the parabola, $x^{2} - 6 x - y + 13 = 0$ , has $y$ -intercept equal to ___

Q: How do I know if the parabola opens up or down?

Great question! Once you rewrite the parabola as $(x - 3)^2 = y - 4$, it takes the form $(x - h)^2 = 4a(y - k)$. Here, $4a = 1$, which means $a = 1/4$. Since $a > 0$, the parabola opens upwards. Visualizing this helps you confirm the vertex position.

Q: Is there a shortcut for the center of the circle $x^2 + y^2 + 2x = 0$?

Absolutely! For any circle in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is always $(-g, -f)$. In your equation, $2g = 2$, so $g = 1$, and $2f = 0$, so $f = 0$. This immediately gives you the center at $(-1, 0)$ without needing to complete the square.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Do not let the complex number notation intimidate you. The equation involves

z = x + i y

. Recall that

z^{2} = (x + i y)^{2} = x^{2} - y^{2} + 2 i x y

. The real part,

Re (z^{2})

, is simply

x^{2} - y^{2}

. Substitute this and the other components into the equation to see the circle emerge.

Hint 2

The Arsenal: Once you have the Cartesian equation

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

, complete the square to find the center. For the parabola

x^{2} - 6 x - y + 13 = 0

, isolate the

y

term and complete the square for

x

to identify the vertex

(h, k)

in the form

(x - h)^{2} = 4 a (y - k)

Hint 3

The Execution: You now have two points: the center of the circle

C (- 1, 0)

and the vertex of the parabola

V (3, 4)

. Use the slope formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

to find the slope of the line connecting them, then use the point-slope form

y - y_{1} = m (x - x_{1})

to find the equation of the line and its

y

-intercept.

Visualized Solution

Introduction to the Problem

Given equation: $Re (z^{2}) + 2 (Im (z))^{2} + 2 Re (z) = 0$
Substitute $z = x + i y$ to convert to Cartesian form.

Expanding $z^{2}$

$z^{2} = (x + i y)^{2} = x^{2} + (i y)^{2} + 2 i x y$
Since $i^{2} = - 1$ , $z^{2} = (x^{2} - y^{2}) + i (2 x y)$
Therefore, $Re (z^{2}) = x^{2} - y^{2}$

Cartesian Form of Circle

Substitute $Re (z^{2}) = x^{2} - y^{2}$ , $Im (z) = y$ , and $Re (z) = x$ :
$(x^{2} - y^{2}) + 2 (y)^{2} + 2 x = 0$
Simplify: $x^{2} + y^{2} + 2 x = 0$

Center of the Circle

Compare $x^{2} + y^{2} + 2 x = 0$ with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ :
$2 g = 2 ⟹ g = 1$
$2 f = 0 ⟹ f = 0$
Center $C = (- g, - f) = (- 1, 0)$

Analyzing the Parabola

Given Parabola: $x^{2} - 6 x - y + 13 = 0$
Goal: Rewrite in the form $(x - h)^{2} = 4 a (y - k)$ to find vertex $(h, k)$ .

Completing the Square

$x^{2} - 6 x + 9 - 9 - y + 13 = 0$
$(x - 3)^{2} - y + 4 = 0$
$(x - 3)^{2} = y - 4$

Vertex of the Parabola

Comparing $(x - 3)^{2} = y - 4$ with $(x - h)^{2} = 4 a (y - k)$ :
Vertex $V = (h, k) = (3, 4)$

Slope of the Connecting Line

Points: $C (- 1, 0)$ and $V (3, 4)$
Slope $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{4 - 0}{3 - ( - 1 )}$
$m = \frac{4}{4} = 1$

Equation of the Line

Using point $(- 1, 0)$ and $m = 1$ :
$y - 0 = 1 (x - (- 1))$
Equation: $y = x + 1$

Calculating the $y$ -intercept

To find the $y$ -intercept, set $x = 0$ in the line equation:
$y = 0 + 1$
$y = 1$

Final Answer

Key Takeaways:
Convert complex equations to Cartesian form using $z = x + i y$ .
Complete the square to find the vertex of a parabola.
The line equation through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $y - y_{1} = m (x - x_{1})$ .
Final Answer: $y$ -intercept = $1$ .

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Complex Geometry

To demystify the complex equation

Re (z^{2}) + 2 (Im (z))^{2} + 2 Re (z) = 0

, we utilize the standard substitution

z = x + i y

Expanding

z^{2}

yields:

z^{2} = (x + i y)^{2} = (x^{2} - y^{2}) + i (2 x y)

The real part,

Re (z^{2})

, is therefore

x^{2} - y^{2}

. Substituting

Im (z) = y

and

Re (z) = x

into the original equation, we obtain:

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

Simplifying this expression leads to the Cartesian equation of a circle:

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

By comparing this to the general form

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

, we identify the center of the circle as

C (- 1, 0)

Unveiling the Parabola

Next, we analyze the parabola defined by

x^{2} - 6 x - y + 13 = 0

. To find its vertex, we convert the equation into the standard vertex form

(x - h)^{2} = 4 a (y - k)

We complete the square for the

x

terms:

x^{2} - 6 x + 9 - 9 - y + 13 = 0

(x - 3)^{2} - y + 4 = 0

This simplifies to the standard form:

(x - 3)^{2} = y - 4

From this form, we identify the vertex of the parabola as

V (3, 4)

The Final Connection

We must now find the

y

-intercept of the line passing through the circle's center

C (- 1, 0)

and the parabola's vertex

V (3, 4)

. First, we calculate the slope

m

of the line:

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{4 - 0}{3 - ( - 1 )} = \frac{4}{4} = 1

Using the point-slope form

y - y_{1} = m (x - x_{1})

with the point

(- 1, 0)

, the equation of the line becomes:

y - 0 = 1 (x - (- 1)) \Rightarrow y = x + 1

To find the

y

-intercept, we set

x = 0

in the line equation. This yields

y = 1

The final

y

-intercept is $1$ .

Common Traps & FAQs

Wait, why do we replace

z

with

x + i y

? Isn't there a property of complex numbers we could use?

While there are many elegant properties of complex numbers, in coordinate geometry problems like this, converting to Cartesian form is the most robust and reliable method. It transforms an abstract algebraic expression into a concrete geometric locus, allowing you to use your standard toolkit for circles and parabolas.

How do I know if the parabola opens up or down?

Great question! Once you rewrite the parabola as

(x - 3)^{2} = y - 4

, it takes the form

(x - h)^{2} = 4 a (y - k)

. Here,

4 a = 1

, which means

a = 1/4

. Since

a > 0

, the parabola opens upwards. Visualizing this helps you confirm the vertex position.

Is there a shortcut for the center of the circle

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

Absolutely! For any circle in the general form

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

, the center is always

(- g, - f)

. In your equation,

2 g = 2

, so

g = 1

, and

2 f = 0

, so

f = 0

. This immediately gives you the center at

(- 1, 0)

without needing to complete the square.

Animated Solution for Mathematics - Circles: The equation of a circle is $Re (z^{2}) + 2 (Im (z))^{2} + 2 Re (z) = 0$ , where $z = x + i y$ . A line which passes through the center of the given circle and the vertex of the parabola, $x^{2} - 6 x - y + 13 = 0$ , has $y$ -intercept equal to ___

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Expanding $z^{2}$

Cartesian Form of Circle

Center of the Circle

Analyzing the Parabola

Completing the Square

Vertex of the Parabola

Slope of the Connecting Line

Equation of the Line

Calculating the $y$ -intercept

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Complex Geometry

Unveiling the Parabola

The Final Connection

Similar Questions

The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

Let a circle $C$ touch the lines $L_{1} : 4 x - 3 y + K_{1} = 0$ and $L_{2} : 4 x - 3 y + K_{2} = 0, K_{1}, K_{2} \in R$ . If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(- 1, 2)$ and $L_{2}$ at $(3, - 6)$ , then the equation of the circle $C$ is

Intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . Equation of the circle on $A B$ as a diameter is

Let $C$ be the circle of minimum area touching the parabola $y = 6 - x^{2}$ and the lines $y = 3 ∣ x ∣$ . Then, which one of the following points lies on the circle $C$ ?

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at $(- 1, 0)$ and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the $y$ -axis at the point $P$ , then the square of the distance of $P$ from the centre of $C_{1}$ is :

The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . Equation of the circle with $A B$ as a diameter is .........

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle is

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to

Let the parabola $y = x^{2} + p x - 3$ , meet the coordinate axes at the points P, Q and R. If the circle C with centre at $(- 1, - 1)$ passes through the points P, Q and R, then the area of $△ P QR$ is:

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2

Cartesian Form of Circle

Center of the Circle

Analyzing the Parabola

Completing the Square

Vertex of the Parabola

Slope of the Connecting Line

Equation of the Line

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

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Complex Geometry

Unveiling the Parabola

The Final Connection

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters o of the circle x2+y2−2x−6y+6=0 is a chord of another circle 'C' whose center is at (2, 1), then its radius is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle passes through the points (2,3) and (4,5). If its centre lies on the line, y−4x+3=0, then its radius is equal to

Expanding $z^{2}$

Calculating the $y$ -intercept

The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

Intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . Equation of the circle on $A B$ as a diameter is

Let $C$ be the circle of minimum area touching the parabola $y = 6 - x^{2}$ and the lines $y = 3 ∣ x ∣$ . Then, which one of the following points lies on the circle $C$ ?

The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . Equation of the circle with $A B$ as a diameter is .........

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle is

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to