JEE Main 2022 (29 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let the mirror image of a circle $c_{1} : x^{2} + y^{2} - 2 x - 6 y + α = 0$ in line $y = x + 1$ be $c_{2} : 5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0$ . If $r$ is the radius of circle $c_{2}$ , then $α + 6 r^{2}$ is equal to ________

Q: Why do we divide by 5 in the equation of $c_2$?

The standard form of a circle equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. The coefficients of $x^2$ and $y^2$ must be 1. The given equation $5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$ has coefficients of 5, so we must divide the entire equation by 5 to correctly identify the center $(-g, -f)$ and the constant term $c$.

Q: Is there a sign error in the mirror image formula?

The formula $\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$ is correct. The $-2$ factor is essential because the mirror image is located on the opposite side of the line at the exact same perpendicular distance as the original point.

Enter Numerical Value:

Visualized Solution

Analyzing Circle $c_{1}$

Circle $c_{1} : x^{2} + y^{2} - 2 x - 6 y + α = 0$
Center $C_{1} = (- \frac{- 2}{2}, - \frac{- 6}{2}) = (1, 3)$
Radius $r_{1} = 1^{2} + 3^{2} - α = 10 - α$

The Mirror Line

Mirror Line $L : y = x + 1 ⟹ x - y + 1 = 0$
Property: Reflection preserves the radius ( $r_{1} = r_{2}$ )
The center $C_{2}$ of circle $c_{2}$ is the mirror image of $C_{1}$ in line $L$

Mirror Image Formula

Mirror image $(x_{2}, y_{2})$ of $(x_{1}, y_{1})$ in $a x + b y + c = 0$ :
$\frac{x _{2} - x _{1}}{a} = \frac{y _{2} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$
Substitute $C_{1} (1, 3)$ and line $1 x - 1 y + 1 = 0$

Calculating Center $C_{2}$

$\frac{x _{2} - 1}{1} = \frac{y _{2} - 3}{- 1} = - 2 \frac{1 ( 1 ) - 1 ( 3 ) + 1}{1 ^{2} + ( - 1 ) ^{2}}$
$\frac{x _{2} - 1}{1} = \frac{y _{2} - 3}{- 1} = - 2 \frac{- 1}{2} = 1$
Solving: $x_{2} - 1 = 1 ⟹ x_{2} = 2$
Solving: $y_{2} - 3 = - 1 ⟹ y_{2} = 2$
Center $C_{2} = (2, 2)$

Normalizing Circle $c_{2}$

Circle $c_{2} : 5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0$
Divide by $5$ to normalize:
$x^{2} + y^{2} + 2 g x + 2 f y + \frac{38}{5} = 0$

Finding $g$ and $f$

Center of $c_{2}$ from standard equation: $(- g, - f)$
Equating with geometric center $C_{2} = (2, 2)$ :
$- g = 2 ⟹ g = - 2$
$- f = 2 ⟹ f = - 2$

Radius of Circle $c_{2}$

Radius $r$ of $c_{2}$ : $r^{2} = g^{2} + f^{2} - c$
$r^{2} = (- 2)^{2} + (- 2)^{2} - \frac{38}{5}$
$r^{2} = 4 + 4 - 7.6 = 8 - 7.6 = 0.4$
$r^{2} = \frac{2}{5}$

Equating the Radii

Since $c_{2}$ is the mirror image of $c_{1}$ , $r_{1} = r$
$r_{1}^{2} = r^{2}$
$10 - α = \frac{2}{5}$

Solving for $α$

$α = 10 - \frac{2}{5}$
$α = \frac{50 - 2}{5} = \frac{48}{5}$

Final Calculation

Calculate $α + 6 r^{2}$ :
$\frac{48}{5} + 6 (\frac{2}{5})$
$= \frac{48}{5} + \frac{12}{5}$
$= \frac{60}{5} = 12$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We begin with the circle

c_{1}

defined by the equation:

x^{2} + y^{2} - 2 x - 6 y + α = 0

To determine the geometric properties of

c_{1}

, we complete the square or use the standard form. The center is identified as

C_{1} = (1, 3)

The radius squared is calculated using the formula

r_{1}^{2} = g^{2} + f^{2} - c

r_{1}^{2} = 1^{2} + 3^{2} - α = 10 - α

The Mirror Transformation

We reflect

c_{1}

across the line

y = x + 1

, which can be rewritten as

x - y + 1 = 0

. Since reflection is an isometry, it preserves the radius of the circle.

Therefore, the radius of the reflected circle

c_{2}

must be identical to the radius of

c_{1}

. This gives us the fundamental constraint:

r_{1}^{2} = r_{2}^{2}

To find the center

C_{2}

of the reflected circle, we use the reflection formula for a point

(x_{1}, y_{1})

across the line

a x + b y + c = 0

\frac{x _{2} - x _{1}}{a} = \frac{y _{2} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

Substituting

C_{1} (1, 3)

and the line

x - y + 1 = 0

into this formula, we calculate the coordinates of the reflected center:

C_{2} = (2, 2)

The Normalization Trap

The equation for

c_{2}

is provided as

5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0

. To analyze this correctly, we must normalize the equation by dividing by

5

x^{2} + y^{2} + 2 g x + 2 f y + 7.6 = 0

Comparing this to the standard form

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

, we identify the center as

(- g, - f) = (2, 2)

. This implies

g = - 2

and

f = - 2

We now calculate the radius squared of

c_{2}

r_{2}^{2} = g^{2} + f^{2} - c = (- 2)^{2} + (- 2)^{2} - 7.6

r_{2}^{2} = 8 - 7.6 = 0.4

Final Calculation

Equating the radii squared from our previous steps, we have

10 - α = 0.4

. Solving for

α

yields:

α = 9.6

We are tasked with finding the value of

α + 6 r_{2}^{2}

. Substituting our derived values:

α + 6 r_{2}^{2} = 9.6 + 6 (0.4)

α + 6 r_{2}^{2} = 9.6 + 2.4 = 12

The final result is 12.

Common Traps & FAQs

Why does the radius remain the same after reflection?

Reflection is an isometry, a transformation that preserves distances. When you reflect a circle across a line, the distance from the center to any point on the circumference remains constant. Thus, the radius is invariant.

Why do we divide by 5 in the equation of

c_{2}

The standard form of a circle equation is

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

. The coefficients of

x^{2}

and

y^{2}

must be 1. The given equation

5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0

has coefficients of 5, so we must divide the entire equation by 5 to correctly identify the center

(- g, - f)

and the constant term

c

Is there a sign error in the mirror image formula?

The formula

\frac{x _{2} - x _{1}}{a} = \frac{y _{2} - y _{1}}{b} = - 2 \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

is correct. The

- 2

factor is essential because the mirror image is located on the opposite side of the line at the exact same perpendicular distance as the original point.

Animated Solution for Mathematics - Circles: Let the mirror image of a circle $c_{1} : x^{2} + y^{2} - 2 x - 6 y + α = 0$ in line $y = x + 1$ be $c_{2} : 5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0$ . If $r$ is the radius of circle $c_{2}$ , then $α + 6 r^{2}$ is equal to ________

Enter Numerical Value:

Visualized Solution

Analyzing Circle $c_{1}$

The Mirror Line

Mirror Image Formula

Calculating Center $C_{2}$

Normalizing Circle $c_{2}$

Finding $g$ and $f$

Radius of Circle $c_{2}$

Equating the Radii

Solving for $α$

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Mirror Transformation

The Normalization Trap

Final Calculation

Similar Questions

Consider a circle $C_{1} : x^{2} + y^{2} - 4 x - 2 y = α - 5$ . Let its mirror image in the line $y = 2 x + 1$ be another circle $C_{2} : 5 x^{2} + 5 y^{2} - 10 f x - 10 g y + 36 = 0$ . Let $r$ be the radius of $C_{2}$ . Then $α + r$ is equal to ________

If one of the diameters of the circle $x^{2} + y^{2} - 22 x - 62 y + 14 = 0$ is a chord of the circle $(x - 22)^{2} + (y - 22)^{2} = r^{2}$ , then the value of $r^{2}$ is equal to ____.

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

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

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________

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 :

Let a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}, k > 0$ , touch the x-axis at $(1, 0)$ . If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is 2, then the value of $h + k + r$ is equal to ____.

Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$ . If $T$ is the circle centred at $(0, y)$ , passing through origin and touching the circle $C$ externally, then the radius of $T$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​

The Mirror Line

Mirror Image Formula

Calculating Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Normalizing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

Radius of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2​

Equating the Radii

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

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Mirror Transformation

The Normalization Trap

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters of the circle x2+y2−22​x−62​y+14=0 is a chord of the circle (x−22​)2+(y−22​)2=r2, then the value of r2 is equal to ____.

.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; } If one of the diameters of the circle x2+y2−2x−6y+6=0 is a chord to the circle with centre (2,1), then the radius of the circle is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3​)x+2(4−6​)y=k+63​+86​,k>0, touch internally at the point P(α,β), then (α+3​)2+(β+6​)2 is equal to ________

Analyzing Circle $c_{1}$

Calculating Center $C_{2}$

Normalizing Circle $c_{2}$

Finding $g$ and $f$

Radius of Circle $c_{2}$

Solving for $α$

If one of the diameters of the circle $x^{2} + y^{2} - 22 x - 62 y + 14 = 0$ is a chord of the circle $(x - 22)^{2} + (y - 22)^{2} = r^{2}$ , then the value of $r^{2}$ is equal to ____.

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

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________