JEE Advanced 1986

LEVELJEE Main

Animated Solution for Mathematics - Circles: Lines $5 x + 12 y - 10 = 0$ and $5 x - 12 y - 40 = 0$ touch a circle $C_{1}$ of diameter 6. If the centre of $C_{1}$ lies in the first quadrant, find the equation of the circle $C_{2}$ which is concentric with $C_{1}$ and cuts intercepts of length 8 on these lines.

Visualized Solution

Visualizing the Setup

Given lines: $L_{1} : 5 x + 12 y - 10 = 0$ and $L_{2} : 5 x - 12 y - 40 = 0$
Circle $C_{1}$ touches both lines.
Diameter of $C_{1} = 6 ⟹$ Radius $r = 3$ .

The Distance Formula

Distance from point $(x_{1}, y_{1})$ to line $a x + b y + c = 0$ is:
$d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
For $C_{1}$ , distance from center $(h, k)$ to $L_{1}$ and $L_{2}$ must be $r = 3$ .

Applying Distance to $L_{1}$

Distance to $L_{1} : 5 x + 12 y - 10 = 0$ :
$\frac{∣5 h + 12 k - 10∣}{5 ^{2} + 1 2 ^{2}} = 3$
$\frac{∣5 h + 12 k - 10∣}{13} = 3 ⟹ ∣5 h + 12 k - 10∣ = 39$

Applying Distance to $L_{2}$

Distance to $L_{2} : 5 x - 12 y - 40 = 0$ :
$\frac{∣5 h - 12 k - 40∣}{5 ^{2} + ( - 12 ) ^{2}} = 3$
$\frac{∣5 h - 12 k - 40∣}{13} = 3 ⟹ ∣5 h - 12 k - 40∣ = 39$

Solving for the Centre

To get $h, k > 0$ (first quadrant), we solve:
1) $5 h + 12 k - 10 = 39 ⟹ 5 h + 12 k = 49$
2) $5 h - 12 k - 40 = - 39 ⟹ 5 h - 12 k = 1$
Adding gives $10 h = 50 ⟹ h = 5$
Subtracting gives $24 k = 48 ⟹ k = 2$
Center of $C_{1}$ is $(5, 2)$ .

Introducing Circle $C_{2}$

$C_{2}$ is concentric with $C_{1}$ , so its center is also $(5, 2)$ .
Let $R$ be the radius of $C_{2}$ .
$C_{2}$ cuts an intercept of length 8 on the lines.

The Intercept Formula

Perpendicular distance from $(5, 2)$ to $L_{1}$ is $p = 3$ .
Length of intercept on a line is $L = 2 R^{2} - p^{2}$ .
We form a right-angled triangle with radius $R$ , distance $p$ , and half-chord.

Setting up the Radius Equation

Given intercept length $= 8$ :
$2 R^{2} - 3^{2} = 8$
$R^{2} - 9 = 4$

Calculating the Radius $R$

Squaring both sides:
$R^{2} - 9 = 16$
$R^{2} = 25$
So, the radius of $C_{2}$ is $R = 5$ .

Equation of Circle $C_{2}$

Standard equation of a circle: $(x - h)^{2} + (y - k)^{2} = R^{2}$
Substitute center $(5, 2)$ and $R^{2} = 25$ :
$(x - 5)^{2} + (y - 2)^{2} = 25$

Expanding the Equation

Expanding the terms:
$x^{2} - 10 x + 25 + y^{2} - 4 y + 4 = 25$
Simplifying by canceling 25 from both sides:
$x^{2} + y^{2} - 10 x - 4 y + 4 = 0$

Key Takeaways

Key Takeaway 1: Distance from center to tangent equals the radius.
Key Takeaway 2: Intercept length $L = 2 R^{2} - p^{2}$ connects concentric circles.
Next Challenge: What if the center was in the second quadrant? How would the signs change?

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two lines,

L_{1} : 5 x + 12 y - 10 = 0

and

L_{2} : 5 x - 12 y - 40 = 0

. A circle

C_{1}

is nestled perfectly between them, touching both lines.

Given the diameter of

C_{1}

6

, we immediately determine the radius

r = 3

The Quest for the Center

To define

C_{1}

, we need its center

(h, k)

. Since the circle touches both lines, the perpendicular distance from

(h, k)

to each line must be exactly

3

We use the distance formula:

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

For

L_{1}

, the distance is:

\frac{∣5 h + 12 k - 10∣}{5 ^{2} + 1 2 ^{2}} = 3 \Rightarrow ∣5 h + 12 k - 10∣ = 39

For

L_{2}

, the distance is:

\frac{∣5 h - 12 k - 40∣}{5 ^{2} + 1 2 ^{2}} = 3 \Rightarrow ∣5 h - 12 k - 40∣ = 39

Since the center lies in the first quadrant, we solve the system

5 h + 12 k = 49

and

5 h - 12 k = 1

. Adding these equations gives

10 h = 50

, so

h = 5

. Substituting back, we find

k = 2

. Our center is

(5, 2)

The Concentric Expansion

Now, we introduce

C_{2}

, a larger circle concentric with

C_{1}

. This means

C_{2}

shares the same center

(5, 2)

We are told

C_{2}

cuts an intercept of length

8

on these lines. The perpendicular distance

p

from the center

(5, 2)

to the lines remains

3

The relationship between the radius

R

C_{2}

, the distance

p

, and the intercept length

L

is given by:

L = 2 R^{2} - p^{2}

Substituting

L = 8

and

p = 3

, we get:

2 R^{2} - 3^{2} = 8 \Rightarrow R^{2} - 9 = 4

Squaring both sides yields

R^{2} - 9 = 16

, which results in

R^{2} = 25

The Final Masterpiece

We have the center

(5, 2)

and

R^{2} = 25

. The equation of

C_{2}

is:

(x - 5)^{2} + (y - 2)^{2} = 25

Expanding this, we get:

x^{2} - 10 x + 25 + y^{2} - 4 y + 4 = 25

The

25

on both sides cancels out, leaving us with the elegant final equation: $x^{2} + y^{2} - 10 x - 4 y + 4 = 0$

Common Traps & FAQs

Wait, why do we have absolute values in the distance formula?

The absolute value is essential because the distance must always be positive. When we remove the absolute value bars, we get two cases (positive and negative), which correspond to the two sides of the line. Since we are in the first quadrant, we choose the signs that satisfy

h, k > 0

Why is the intercept length formula

2 R^{2} - p^{2}

Imagine a right-angled triangle where the hypotenuse is the radius

R

, one leg is the perpendicular distance

p

from the center to the chord, and the other leg is half the chord length. By Pythagoras' theorem,

R^{2} = p^{2} + (half-chord)^{2}

. Thus, the half-chord is

R^{2} - p^{2}

, and the full chord is twice that.

Why can't we just use energy conservation?

Energy conservation is for physics! Here, we are in the realm of pure geometry. We rely on the properties of circles and lines. The 'energy' of this problem lies in the elegance of the algebraic cancellation at the end.

Visualized Solution

Visualizing the Setup

The Distance Formula

Applying Distance to $L_{1}$

Applying Distance to $L_{2}$

Solving for the Centre

Introducing Circle $C_{2}$

The Intercept Formula

Setting up the Radius Equation

Calculating the Radius $R$

Equation of Circle $C_{2}$

Expanding the Equation

Key Takeaways

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Quest for the Center

The Concentric Expansion

The Final Masterpiece

Similar Questions

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Visualized Solution

Visualizing the Setup

The Distance Formula

Applying Distance to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Applying Distance to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Solving for the Centre

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

The Intercept Formula

Setting up the Radius Equation

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

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

Expanding the Equation

Key Takeaways

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Quest for the Center

The Concentric Expansion

The Final Masterpiece

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 at the point (5,5).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The points of intersection of the line 4x−3y−10=0 and the circle x2+y2−2x+4y−20=0 are ......... and .........

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 3x−4y−7=0 and 2x−3y−5=0 are two diameters of a circle of area 49π square units, the equation of the circle is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centre of a circle, which touches externally the circle x2+y2−6x−8y+14=0 and also touches the y-axis, is given by the equation:

Applying Distance to $L_{1}$

Applying Distance to $L_{2}$

Introducing Circle $C_{2}$

Calculating the Radius $R$

Equation of Circle $C_{2}$

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

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The points of intersection of the line $4 x - 3 y - 10 = 0$ and the circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ are ......... and .........

Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

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