Two tangents are drawn from a point $P$ to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$, such that the angle between these tangents is $\tan^{-1}(\frac{12}{5})$, where $\tan^{-1}(\frac{12}{5}) \in (0, \pi)$. If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\Delta PAB$ and $\Delta CAB$ is : Options: (A) 11: 4 (B) 9: 4 (C) 3: 1 (D) 2: 1

### Analyzing the Setup Imagine standing at an external point $P$, looking at a circle. You draw two lines, $PA$ and $PB$, that kiss the circle at points $A$ and $B$. This is not just a drawing; it is a dance of symmetry. The problem asks us to find the ratio of the areas of two triangles: $\Delta PAB$ and $\Delta CAB$. ### Unmasking the Circle First, we must understand our circle. The equation $x^2 + y^2 - 2x - 4y + 4 = 0$ is a bit disguised. By completing the square, we rewrite it as: $$(x - 1)^2 + (y - 2)^2 = 1$$ Now, the soul of the circle is revealed: a center $C$ at $(1, 2)$ and a radius $r = 1$. This is our anchor. ### The Trigonometric Bridge We are told the angle between the tangents is $\tan^{-1}(\frac{12}{5})$. Let this total angle be $2\theta$. The line $CP$ acts as a mirror, bisecting this angle, creating two congruent right-angled triangles, $\Delta PAC$ and $\Delta PBC$. In $\Delta PAC$, the angle at $A$ is $90^{\circ}$, and the angle at $P$ is $\theta$. Thus, $\tan\theta = \frac{AC}{PA} = \frac{r}{L}$, where $L$ is the length of the tangent. We know $\tan(2\theta) = \frac{12}{5}$. Using the double-angle identity: $$\frac{2\tan\theta}{1 - \tan^2\theta} = \frac{12}{5}$$ This leads us to the quadratic $6\tan^2\theta + 5\tan\theta - 6 = 0$. Solving this, we find $\tan\theta = \frac{2}{3}$. Since $\tan\theta = \frac{r}{L}$ and $r = 1$, we immediately find $L = \frac{3}{2}$. ### The Area Ratio: A Beautiful Cancellation Now, for the climax. The area of $\Delta PAB$ is given by the formula: $$\text{Area}(\Delta PAB) = \frac{r L^3}{r^2 + L^2}$$ The area of $\Delta CAB$ is given by: $$\text{Area}(\Delta CAB) = \frac{r^3 L}{r^2 + L^2}$$ When we take the ratio $\frac{\text{Area } \Delta PAB}{\text{Area } \Delta CAB}$, the denominators $r^2 + L^2$ vanish into thin air! We are left with: $$\frac{r L^3}{r^3 L} = \frac{L^2}{r^2}$$ Substituting our values, $L = \frac{3}{2}$ and $r = 1$, the ratio becomes: $$\left(\frac{3}{2}\right)^2 / 1^2 = \frac{9}{4}$$ The final ratio is **$9:4$**. This result is not just a number; it is a testament to the power of geometric insight over brute-force calculation.

Why do we reject the negative root for $\tan\theta$?

In our right-angled triangle $\Delta PAC$, $\theta$ is one of the interior angles. Since the sum of angles in a triangle is $180^\circ$ and one angle is $90^\circ$, $\theta$ must be an acute angle (between $0$ and $90^\circ$). The tangent of an acute angle is always positive, so we must reject any negative values obtained from the quadratic equation.

JEE Main 2021 (17 March Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: Two tangents are drawn from a point $P$ to the circle $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ , such that the angle between these tangents is $tan^{- 1} (\frac{12}{5})$ , where $tan^{- 1} (\frac{12}{5}) \in (0, π)$ . If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$ , then the ratio of the areas of $Δ P A B$ and $Δ C A B$ is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Circle: $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$
Point $P$ is the external point from which tangents $P A$ and $P B$ are drawn.
Angle between tangents $∠ A P B = 2 θ = tan^{- 1} (\frac{12}{5})$ .

Standard Form of the Circle

Rearranging the equation: $(x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) = 1$
Standard form: $(x - 1)^{2} + (y - 2)^{2} = 1$
Center $C = (1, 2)$ and Radius $r = 1$

Defining the Angle $θ$

Let the total angle between tangents be $2 θ$ .
Given: $2 θ = tan^{- 1} (\frac{12}{5}) ⟹ tan (2 θ) = \frac{12}{5}$
The line $C P$ bisects the angle, so $∠ A P C = θ$ .

Using the Double Angle Formula

We know $tan (2 θ) = \frac{12}{5}$
Using the identity: $\frac{2 t a n θ}{1 - t a n ^{2} θ} = \frac{12}{5}$
Cross-multiplying: $10 tan θ = 12 - 12 tan^{2} θ$

Solving for $tan θ$

Rearranging into a standard quadratic: $6 tan^{2} θ + 5 tan θ - 6 = 0$
Factoring: $(2 tan θ + 3) (3 tan θ - 2) = 0$
Since $θ$ is acute, $tan θ$ must be positive.
Therefore, $tan θ = \frac{2}{3}$

Finding the Tangent Length $L$

In right $Δ P A C$ , $tan θ = \frac{Opposite}{Adjacent} = \frac{A C}{P A}$
Here, $A C = r$ and $P A = L$ (Length of tangent)
Substitute values: $\frac{2}{3} = \frac{1}{L}$
Tangent Length $L = \frac{3}{2}$

Area of $Δ P A B$

The chord of contact is $A B$ .
Area of $Δ P A B = \frac{r L ^{3}}{r ^{2} + L ^{2}}$
This is the triangle formed by the two tangents and the chord $A B$ .

Area of $Δ C A B$

Area of $Δ C A B = \frac{r ^{3} L}{r ^{2} + L ^{2}}$
This is the triangle formed by the two radii and the chord $A B$ .

Calculating the Ratio

We need the ratio: $\frac{Area Δ P A B}{Area Δ C A B}$
Ratio $= \frac{\frac{r L ^{3}}{r ^{2} + L ^{2}}}{\frac{r ^{3} L}{r ^{2} + L ^{2}}}$
The denominators cancel out.
Simplifying the ratio: $\frac{L ^{2}}{r ^{2}}$

Final Answer

Substitute $L = \frac{3}{2}$ and $r = 1$ :
Ratio $= \frac{( 3/2 ) ^{2}}{1 ^{2}} = \frac{9/4}{1}$
The ratio of the areas is $9 : 4$ .

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine standing at an external point

P

, looking at a circle. You draw two lines,

P A

and

P B

, that kiss the circle at points

A

and

B

This is not just a drawing; it is a dance of symmetry. The problem asks us to find the ratio of the areas of two triangles:

Δ P A B

and

Δ C A B

Unmasking the Circle

First, we must understand our circle. The equation

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

is a bit disguised.

By completing the square, we rewrite it as:

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

Now, the soul of the circle is revealed: a center

C

(1, 2)

and a radius

r = 1

. This is our anchor.

The Trigonometric Bridge

We are told the angle between the tangents is

tan^{- 1} (\frac{12}{5})

. Let this total angle be

2 θ

The line

C P

acts as a mirror, bisecting this angle, creating two congruent right-angled triangles,

Δ P A C

and

Δ P B C

. In

Δ P A C

, the angle at

A

9 0^{\circ}

, and the angle at

P

θ

Thus,

tan θ = \frac{A C}{P A} = \frac{r}{L}

, where

L

is the length of the tangent. We know

tan (2 θ) = \frac{12}{5}

Using the double-angle identity:

\frac{2 tan θ}{1 - tan ^{2} θ} = \frac{12}{5}

This leads us to the quadratic

6 tan^{2} θ + 5 tan θ - 6 = 0

. Solving this, we find

tan θ = \frac{2}{3}

Since

tan θ = \frac{r}{L}

and

r = 1

, we immediately find

L = \frac{3}{2}

The Area Ratio

A Beautiful Cancellation

Now, for the climax. The area of

Δ P A B

is given by the formula:

Area (Δ P A B) = \frac{r L ^{3}}{r ^{2} + L ^{2}}

The area of

Δ C A B

is given by:

Area (Δ C A B) = \frac{r ^{3} L}{r ^{2} + L ^{2}}

When we take the ratio

\frac{Area Δ P A B}{Area Δ C A B}

, the denominators

r^{2} + L^{2}

vanish into thin air! We are left with:

\frac{r L ^{3}}{r ^{3} L} = \frac{L ^{2}}{r ^{2}}

Substituting our values,

L = \frac{3}{2}

and

r = 1

, the ratio becomes:

(\frac{3}{2})^{2} / 1^{2} = \frac{9}{4}

The final ratio is $9 : 4$ . This result is not just a number; it is a testament to the power of geometric insight over brute-force calculation.

Common Traps & FAQs

Why does the line

C P

bisect the angle between the tangents?

This is a fundamental property of circles. The two triangles

Δ P A C

and

Δ P B C

are congruent by the RHS (Right-angle, Hypotenuse, Side) congruence criterion, as both have a right angle at the point of tangency, share the hypotenuse

C P

, and have equal radii

A C = B C = r

. Thus, the angles

∠ A P C

and

∠ B P C

must be equal, meaning

C P

bisects the total angle

∠ A P B

Is there a simpler way to find the area ratio without calculating coordinates?

Absolutely! In fact, calculating coordinates is unnecessary here. By using the geometric properties of the triangles

Δ P A B

and

Δ C A B

, we can express their areas purely in terms of the radius

r

and the tangent length

L

. The ratio simplifies beautifully, making the coordinate calculation redundant.

Why do we reject the negative root for

tan θ

In our right-angled triangle

Δ P A C

θ

is one of the interior angles. Since the sum of angles in a triangle is

18 0^{\circ}

and one angle is

9 0^{\circ}

θ

must be an acute angle (between

0

and

9 0^{\circ}

). The tangent of an acute angle is always positive, so we must reject any negative values obtained from the quadratic equation.

Select Answer:

Visualized Solution

Visualizing the Geometry

Standard Form of the Circle

Defining the Angle $θ$

Using the Double Angle Formula

Solving for $tan θ$

Finding the Tangent Length $L$

Area of $Δ P A B$

Area of $Δ C A B$

Calculating the Ratio

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unmasking the Circle

The Trigonometric Bridge

The Area Ratio

Similar Questions

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

Points $A, B$ and $C$ lie on the parabola $y^{2} = 4 a x$ . The tangents to the parabola at $A, B$ and $C$ , taken in pairs, intersect at points $P, Q$ and $R$ . Determine the ratio of the areas of the triangles $A B C$ and $P QR$ .

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

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Let $A B$ be a chord of length 12 of the circle $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________

Select Answer:

Visualized Solution

Visualizing the Geometry

Standard Form of the Circle

Defining the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Using the Double Angle Formula

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

Finding the Tangent Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

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

Calculating the Ratio

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unmasking the Circle

The Trigonometric Bridge

The Area Ratio

Similar Questions

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Solving for $tan θ$

Finding the Tangent Length $L$

Area of $Δ P A B$

Area of $Δ C A B$

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