The angle between a pair of tangents drawn from a point $P$ to the circle $x^2 + y^2 + 4x - 6y + 9 \sin^2 \alpha + 13 \cos^2 \alpha = 0$ is $2\alpha$. The equation of the locus of the point $P$ is Options: (A) $x^2 + y^2 + 4x - 6y + 4 = 0$ (B) $x^2 + y^2 + 4x - 6y - 9 = 0$ (C) $x^2 + y^2 + 4x - 6y - 4 = 0$ (D) $x^2 + y^2 + 4x - 6y + 9 = 0$

### Analyzing the Setup The given equation is $x^2 + y^2 + 4x - 6y + 9 \sin^2 \alpha + 13 \cos^2 \alpha = 0$. This is a circle equation disguised by trigonometric parameters. To reveal its structure, we group the $x$ and $y$ terms and complete the square: $$(x^2 + 4x) + (y^2 - 6y) = -9 \sin^2 \alpha - 13 \cos^2 \alpha$$ Adding $4$ and $9$ to both sides, we obtain: $$(x+2)^2 + (y-3)^2 = 13 - 9 \sin^2 \alpha - 13 \cos^2 \alpha$$ ### Simplifying the Radius We simplify the right-hand side using the identity $\cos^2 \alpha = 1 - \sin^2 \alpha$: $$13 - 9 \sin^2 \alpha - 13(1 - \sin^2 \alpha) = 13 - 9 \sin^2 \alpha - 13 + 13 \sin^2 \alpha = 4 \sin^2 \alpha$$ The equation of the circle becomes: $$(x+2)^2 + (y-3)^2 = (2 \sin \alpha)^2$$ Thus, the circle has a center $C(-2, 3)$ and a radius $R = 2 \sin \alpha$. ### The Geometric Insight Consider a point $P(h, k)$ outside the circle. If the angle between the two tangents drawn from $P$ to the circle is $2\alpha$, the line $CP$ bisects this angle. By dropping a perpendicular from $C$ to the point of tangency, we form a right-angled triangle. In this triangle, the angle at $P$ is $\alpha$, the opposite side is the radius $R$, and the hypotenuse is the distance $CP$. Using trigonometry, we establish the relationship: $$\sin \alpha = \frac{R}{CP}$$ ### The Elegant Cancellation Substituting $R = 2 \sin \alpha$ into the trigonometric relation, we get: $$\sin \alpha = \frac{2 \sin \alpha}{CP}$$ The $\sin \alpha$ terms cancel out, yielding $CP = 2$. This implies that the distance from the moving point $P$ to the fixed center $C$ is constant. The locus of $P$ is a circle centered at $C(-2, 3)$ with a radius of $2$. The equation is: $$(x+2)^2 + (y-3)^2 = 2^2$$ Expanding this, we arrive at the final result: **$x^2 + y^2 + 4x - 6y + 9 = 0$**

LEVELJEE Main

Animated Solution for Mathematics - Circles: The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

Q: Can I use the formula for the angle between tangents directly instead of drawing the triangle?

You absolutely can! The formula for the angle $2\theta$ between tangents from a point $(x_1, y_1)$ to a circle $x^2 + y^2 = r^2$ is $\tan \theta = \frac{r}{\sqrt{d^2 - r^2}}$, where $d$ is the distance from the center. However, deriving it from the right-angled triangle as we did is often safer and helps you visualize the geometry, which is a vital skill for JEE Advanced problems.

Q: What happens to the $\alpha$ in the final equation? Is it supposed to disappear?

Yes, it is! The problem asks for the locus of point $P$ such that the angle between the tangents is $2\alpha$. Since the locus is a geometric path that $P$ must follow, it should be independent of the specific parameter $\alpha$ used to define the condition. The fact that $\alpha$ cancels out is a sign that your derivation is correct and that the locus is a fixed, well-defined circle.

Select Answer:

Progressive Hints

Hint 1

The circle equation looks intimidating with its trigonometric coefficients, but do not panic. Your first task is to complete the square for both

x

and

y

terms. Once you isolate the constant on the right side, use the identity

cos^{2} α = 1 - sin^{2} α

to simplify the radius squared.

Hint 2

Visualize the geometry of the problem. You have a point

P

outside the circle. When you draw two tangents from

P

, the line segment connecting the center

C

P

acts as an angle bisector. This creates a right-angled triangle where the angle at

P

α

and the opposite side is the radius

R

Hint 3

Use the trigonometric relationship in the right-angled triangle formed by the center, the point of tangency, and point

P

. Specifically,

sin α = \frac{R}{C P}

. Since you have already found

R

in terms of

α

, you will see that

sin α

cancels out, leaving you with a constant distance

C P

. The locus is simply a circle centered at

C

with radius

C P

Visualized Solution

The Given Circle Equation

Given circle: $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$

Converting to Standard Form

Standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$
Grouping terms: $(x^{2} + 4 x) + (y^{2} - 6 y) = - 9 sin^{2} α - 13 cos^{2} α$

Completing the Square

Add $4$ for $x$ and $9$ for $y$ on both sides.
$(x^{2} + 4 x + 4) + (y^{2} - 6 y + 9) = 13 - (9 sin^{2} α + 13 cos^{2} α)$

Simplifying the Constant

Right Hand Side: $13 - 9 sin^{2} α - 13 cos^{2} α$
Use identity: $cos^{2} α = 1 - sin^{2} α$
$13 - 9 sin^{2} α - 13 (1 - sin^{2} α)$

Finding the Radius

$13 - 9 sin^{2} α - 13 + 13 sin^{2} α = 4 sin^{2} α$
Equation: $(x + 2)^{2} + (y - 3)^{2} = (2 sin α)^{2}$
Center $C (- 2, 3)$ and Radius $R = 2 sin α$

The Point $P$ and Tangents

Let the point be $P (h, k)$ .
Angle between the two tangents from $P$ is $2 α$ .

Geometry of the Tangents

Join center $C$ to point $P$ .
The line $C P$ bisects the angle between the tangents.
Angle between $C P$ and a tangent is $α$ .

The Right-Angled Triangle

Draw the radius to the point of tangency.
The radius is perpendicular to the tangent.
In right triangle: $sin α = \frac{R}{C P}$

Calculating Distance $C P$

Substitute $R = 2 sin α$ :
$sin α = \frac{2 s i n α}{C P}$
$C P = \frac{2 s i n α}{s i n α} = 2$

Setting Up the Locus Equation

The distance from $P (h, k)$ to $C (- 2, 3)$ is always $2$ .
Distance formula: $(h - (- 2))^{2} + (k - 3)^{2} = 2$

Expanding the Equation

Squaring both sides: $(h + 2)^{2} + (k - 3)^{2} = 4$
Expanding: $h^{2} + 4 h + 4 + k^{2} - 6 k + 9 = 4$

Final Equation of the Locus

Simplifying: $h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
Replace $(h, k)$ with $(x, y)$ :
$x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given equation is

x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0

. This is a circle equation disguised by trigonometric parameters.

To reveal its structure, we group the

x

and

y

terms and complete the square:

(x^{2} + 4 x) + (y^{2} - 6 y) = - 9 sin^{2} α - 13 cos^{2} α

Adding

4

and

9

to both sides, we obtain:

(x + 2)^{2} + (y - 3)^{2} = 13 - 9 sin^{2} α - 13 cos^{2} α

Simplifying the Radius

We simplify the right-hand side using the identity

cos^{2} α = 1 - sin^{2} α

13 - 9 sin^{2} α - 13 (1 - sin^{2} α) = 13 - 9 sin^{2} α - 13 + 13 sin^{2} α = 4 sin^{2} α

The equation of the circle becomes:

(x + 2)^{2} + (y - 3)^{2} = (2 sin α)^{2}

Thus, the circle has a center

C (- 2, 3)

and a radius

R = 2 sin α

The Geometric Insight

Consider a point

P (h, k)

outside the circle. If the angle between the two tangents drawn from

P

to the circle is

2 α

, the line

C P

bisects this angle.

By dropping a perpendicular from

C

to the point of tangency, we form a right-angled triangle. In this triangle, the angle at

P

α

, the opposite side is the radius

R

, and the hypotenuse is the distance

C P

Using trigonometry, we establish the relationship:

sin α = \frac{R}{C P}

The Elegant Cancellation

Substituting

R = 2 sin α

into the trigonometric relation, we get:

sin α = \frac{2 sin α}{C P}

The

sin α

terms cancel out, yielding

C P = 2

. This implies that the distance from the moving point

P

to the fixed center

C

is constant.

The locus of

P

is a circle centered at

C (- 2, 3)

with a radius of

2

. The equation is:

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

Expanding this, we arrive at the final result: $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Common Traps & FAQs

Wait, why does the line from the center to the external point bisect the angle between the tangents?

This is a beautiful property of circle geometry! If you draw the two tangents from an external point

P

to a circle with center

C

, and connect

C

to the points of tangency, you form two right-angled triangles. These triangles are congruent by the RHS (Right-angle, Hypotenuse, Side) criterion because they share the hypotenuse

C P

and have equal radii. Because the triangles are congruent, the angles at

P

must be equal, meaning the line

C P

bisects the total angle

2 α

into two angles of

α

each.

Can I use the formula for the angle between tangents directly instead of drawing the triangle?

You absolutely can! The formula for the angle

2 θ

between tangents from a point

(x_{1}, y_{1})

to a circle

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

tan θ = \frac{r}{d ^{2} - r ^{2}}

, where

d

is the distance from the center. However, deriving it from the right-angled triangle as we did is often safer and helps you visualize the geometry, which is a vital skill for JEE Advanced problems.

What happens to the

α

in the final equation? Is it supposed to disappear?

Yes, it is! The problem asks for the locus of point

P

such that the angle between the tangents is

2 α

. Since the locus is a geometric path that

P

must follow, it should be independent of the specific parameter

α

used to define the condition. The fact that

α

cancels out is a sign that your derivation is correct and that the locus is a fixed, well-defined circle.

Animated Solution for Mathematics - Circles: The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

Select Answer:

Visualized Solution

The Given Circle Equation

Converting to Standard Form

Completing the Square

Simplifying the Constant

Finding the Radius

The Point $P$ and Tangents

Geometry of the Tangents

The Right-Angled Triangle

Calculating Distance $C P$

Setting Up the Locus Equation

Expanding the Equation

Final Equation of the Locus

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Simplifying the Radius

The Geometric Insight

The Elegant Cancellation

Similar Questions

Let $P Q$ and $M N$ be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $∠ A O B = π /3$ . Then the locus of the point of intersection of the lines $P Q$ and $M N$ is :

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the x-axis is

If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ :

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

Select Answer:

Visualized Solution

The Given Circle Equation

Converting to Standard Form

Completing the Square

Simplifying the Constant

Finding the Radius

The Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Tangents

Geometry of the Tangents

The Right-Angled Triangle

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

Setting Up the Locus Equation

Expanding the Equation

Final Equation of the Locus

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Simplifying the Radius

The Geometric Insight

The Elegant Cancellation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A possible equation of L is

A common tangent of the two circles is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the parabola y2=4ax is 45∘. Show that the locus of the point P is a hyperbola.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coordinates of A and B are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocenter of the triangle PAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the locus of the point whose distances from the point P and the line AB are equal, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=16(x−3) are at right angles, then the locus of point P is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the common tangent touching the circle (x−3)2+y2=9 and the parabola y2=4x above the x-axis is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=4x are at right angles, then the locus of P is

The Point $P$ and Tangents

Calculating Distance $C P$

A possible equation of $L$ is

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the x-axis is

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is