JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: The tangent and the normal lines at the point $(3, 1)$ to the circle $x^{2} + y^{2} = 4$ and the x-axis form a triangle. The area of this triangle (in square units) is :

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry: You are dealing with a circle centered at the origin. The triangle is formed by the tangent line, the normal line, and the x-axis. Can you identify the three vertices of this triangle? One is the point of contact, one is the origin, and the third is where the tangent hits the x-axis.

Hint 2

Recall your toolkit: For a circle

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

, the tangent at

(x_{1}, y_{1})

is given by

x x_{1} + y y_{1} = r^{2}

. Also, remember the fundamental property of the normal to a circle: it always passes through the center of the circle. This simplifies your life immensely!

Hint 3

The final push: Once you have the equations of the tangent and the normal, find their x-intercepts. The area of the triangle is

\frac{1}{2} \times base \times height

. The base is the segment on the x-axis between the two intercepts, and the height is simply the y-coordinate of the point of contact.

Visualized Solution

The Circle and Point $P$

Circle Equation: $x^{2} + y^{2} = 4$
Center: $(0, 0)$ , Radius: $r = 2$
Given Point: $P (3, 1)$
Verification: $(3)^{2} + (1)^{2} = 3 + 1 = 4$

Equation of the Tangent Line

Tangent to $x^{2} + y^{2} = r^{2}$ at $(x_{1}, y_{1})$
Formula: $x x_{1} + y y_{1} = r^{2}$

Substituting Values for Tangent

Substitute $x_{1} = 3$ , $y_{1} = 1$ , $r^{2} = 4$
$3 x + (1) y = 4$
Equation: $3 x + y = 4$

Finding the X-intercept of Tangent

The tangent intersects the x-axis at point $A$ .
Set $y = 0$ in $3 x + y = 4$
$3 x + 0 = 4 \Rightarrow x = \frac{4}{3}$
Point $A = (\frac{4}{3}, 0)$

The Normal Line Property

Normal line is perpendicular to the tangent at point $P$ .
Property: Normal to a circle always passes through its center $(0, 0)$ .

Equation of the Normal Line

Passes through $(0, 0)$ and $P (3, 1)$
Slope $m = \frac{1 - 0}{3 - 0} = \frac{1}{3}$
Equation: $y = \frac{1}{3} x$

Finding the X-intercept of Normal

The normal intersects the x-axis at point $B$ .
Set $y = 0$ in $y = \frac{1}{3} x$
$0 = \frac{1}{3} x \Rightarrow x = 0$
Point $B = (0, 0)$

Identifying the Triangle

Triangle is formed by Tangent, Normal, and X-axis.
Vertices: $P (3, 1)$ , $A (\frac{4}{3}, 0)$ , $B (0, 0)$

Area Formula Setup

Area of $△ P A B = \frac{1}{2} \times Base \times Height$
Base is the segment $A B$ on the x-axis.
Height is the perpendicular distance from $P$ to the x-axis.

Calculating Base and Height

Base $A B = x_{A} - x_{B} = \frac{4}{3} - 0 = \frac{4}{3}$
Height $h = y$ -coordinate of $P = 1$

Final Area Calculation

Area $= \frac{1}{2} \times \frac{4}{3} \times 1$
Area $= \frac{2}{3}$ square units

Conclusion

Final Answer: The area of the triangle is $\frac{2}{3}$ sq. units.
Pro Tip: Using standard tangent formulas ( $x x_{1} + y y_{1} = r^{2}$ ) and normal properties saves crucial time in JEE.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the equation of the circle:

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

This represents a circle centered at the origin

(0, 0)

with a radius

r = 2

. We are given a point

P (3, 1)

Before proceeding, we verify that this point lies on the circle. Substituting the coordinates into the equation:

(3)^{2} + (1)^{2} = 3 + 1 = 4

Since the equation holds true, point

P

lies exactly on the boundary of the circle.

The Tangent Line Equation

To find the tangent line at

P

, we utilize the standard formula for a circle

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

. The equation of the tangent at point

(x_{1}, y_{1})

is given by:

x x_{1} + y y_{1} = r^{2}

Substituting our known values

x_{1} = 3

y_{1} = 1

, and

r^{2} = 4

, we obtain:

3 x + y = 4

To find the x-intercept of this tangent line, we set

y = 0

3 x = 4 \Rightarrow x = \frac{4}{3}

Thus, the tangent intersects the x-axis at point

A (\frac{4}{3}, 0)

The Normal Line

The normal line is perpendicular to the tangent at the point of contact. Crucially, the normal to any circle always passes through its center.

Since the center is

(0, 0)

and the point of contact is

P (3, 1)

, the normal line passes through the origin. Its equation is:

y = (\frac{1 - 0}{3 - 0}) x \Rightarrow y = \frac{1}{3} x

Setting

y = 0

to find the x-intercept of the normal, we find

x = 0

. Therefore, the normal intersects the x-axis at point

B (0, 0)

, which is the origin.

Final Area Calculation

We now have a triangle

P A B

with vertices at

P (3, 1)

A (\frac{4}{3}, 0)

, and

B (0, 0)

The base

A B

lies along the x-axis. Its length is the distance between the origin and

x = \frac{4}{3}

Base = \frac{4}{3}

The height of the triangle is the perpendicular distance from

P

to the x-axis, which is simply the y-coordinate of

P

Height = 1

Using the area formula

Area = \frac{1}{2} \times base \times height

, we calculate:

Area = \frac{1}{2} \times (\frac{4}{3}) \times 1 = \frac{2}{3}

The final area of the triangle is $\frac{2}{3}$ square units.

Common Traps & FAQs

Why does the normal line always pass through the center? Can't I just find the slope of the tangent and take the negative reciprocal?

You absolutely can! Calculating the slope of the tangent and then finding the negative reciprocal to get the normal slope is mathematically correct. However, in the JEE, time is your most precious currency. The geometric property that the normal to a circle is always a radial line passing through the center is a powerful shortcut. It saves you from tedious slope calculations and potential sign errors. Always look for these geometric shortcuts!

Wait, why is the height of the triangle just the y-coordinate of point P?

Think about the geometry. The base of the triangle lies on the x-axis. The height is the perpendicular distance from the vertex

P

to the base. Since the base is on the x-axis (where

y = 0

), the perpendicular distance is simply the vertical displacement of

P

from the x-axis. That vertical displacement is exactly the y-coordinate of

P

Is it possible to solve this using determinants for the area of a triangle?

Yes, absolutely! If you have the coordinates of the three vertices, you can use the determinant formula

\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣

. However, for this specific problem, the base-height method is much faster because the base lies on the x-axis, making the calculation trivial. Always choose the path of least resistance!

Animated Solution for Mathematics - Circles: The tangent and the normal lines at the point $(3, 1)$ to the circle $x^{2} + y^{2} = 4$ and the x-axis form a triangle. The area of this triangle (in square units) is :

Select Answer:

Visualized Solution

The Circle and Point $P$

Equation of the Tangent Line

Substituting Values for Tangent

Finding the X-intercept of Tangent

The Normal Line Property

Equation of the Normal Line

Finding the X-intercept of Normal

Identifying the Triangle

Area Formula Setup

Calculating Base and Height

Final Area Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Line Equation

The Normal Line

Final Area Calculation

Similar Questions

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

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

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

The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the x-axis is :-

The tangent and normal to the ellipse $3 x^{2} + 5 y^{2} = 32$ at the point $P (2, 2)$ meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

If a tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the tangents at the extremities of its major axis at $B$ and $C$ , then the circle with $B C$ as diameter passes through the point:

Select Answer:

Visualized Solution

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

Equation of the Tangent Line

Substituting Values for Tangent

Finding the X-intercept of Tangent

The Normal Line Property

Equation of the Normal Line

Finding the X-intercept of Normal

Identifying the Triangle

Area Formula Setup

Calculating Base and Height

Final Area Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Line Equation

The Normal Line

Final Area Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle x2+y2=4 at (1, rac{\sqrt{3}}{1}) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the tangents at two points A and B on the circle x2+y2−4x+3=0 meet at origin O(0,0). Then the area of the triangle of OAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the smaller of the two circles that touch the parabola, y2=4x at the point (1,2) and the x-axis is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent and normal to the ellipse 3x2+5y2=32 at the point P(2,2) meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

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; } If a tangent to the ellipse x2+4y2=4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

The Circle and Point $P$

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

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

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

The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the x-axis is :-

The tangent and normal to the ellipse $3 x^{2} + 5 y^{2} = 32$ at the point $P (2, 2)$ meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

A possible equation of $L$ is

If a tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the tangents at the extremities of its major axis at $B$ and $C$ , then the circle with $B C$ as diameter passes through the point: