JEE Main 2021 (February)

LEVELJEE Main

Animated Solution for Mathematics - Circles: If the area of the triangle formed by the positive x -axis, the normal and the tangent to the circle $(x - 2)^{2} + (y - 3)^{2} = 25$ at the point (5,7) is A, then 24 A is equal to Note: NTA has dropped this question in the final official answer key.

Enter Numerical Value:

Visualized Solution

Given Circle and Point $P (5, 7)$

Given circle: $(x - 2)^{2} + (y - 3)^{2} = 25$
Point of tangency: $P (5, 7)$

Center and Radius

Standard form: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Center $A (h, k) = (2, 3)$
Radius $r = 25 = 5$

Slope of Normal $m_{N}$

Normal passes through $A (2, 3)$ and $P (5, 7)$
Slope $m_{N} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{7 - 3}{5 - 2} = \frac{4}{3}$

Equation of Normal

Equation: $y - 7 = \frac{4}{3} (x - 5)$
Simplifying: $3 y - 21 = 4 x - 20$
General form: $4 x - 3 y + 1 = 0$

Normal's X-intercept $M$

For x-intercept, set $y = 0$
$4 x - 3 (0) + 1 = 0 \Rightarrow 4 x = - 1$
Point $M = (- \frac{1}{4}, 0)$

Slope of Tangent $m_{T}$

Tangent $⊥$ Normal at $P$
$m_{T} = - \frac{1}{m _{N}} = - \frac{1}{4/3} = - \frac{3}{4}$

Equation of Tangent

Equation: $y - 7 = - \frac{3}{4} (x - 5)$
Simplifying: $4 y - 28 = - 3 x + 15$
General form: $3 x + 4 y - 43 = 0$

Tangent's X-intercept $N$

For x-intercept, set $y = 0$
$3 x + 4 (0) = 43 \Rightarrow x = \frac{43}{3}$
Point $N = (\frac{43}{3}, 0)$

Visualizing Triangle $P M N$

Triangle vertices: $P (5, 7)$ , $M (- \frac{1}{4}, 0)$ , $N (\frac{43}{3}, 0)$
Base $M N$ lies on the x-axis.

Length of Base $M N$

Base $b = x_{N} - x_{M}$
$b = \frac{43}{3} - (- \frac{1}{4}) = \frac{43}{3} + \frac{1}{4}$
$b = \frac{172 + 3}{12} = \frac{175}{12}$

Height of Triangle $h$

Height $h = ∣ y_{P} - 0∣$
$h = 7$

Area of Triangle $A$

Area $A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}$

Final Value of $24 A$

$24 A = 24 \times \frac{1225}{24}$
$24 A = 1225$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the circle equation

(x - 2)^{2} + (y - 3)^{2} = 25

. By comparing this to the standard form

(x - h)^{2} + (y - k)^{2} = r^{2}

, we identify the center

A

(2, 3)

and the radius

r = 5

We are focusing on the point

P (5, 7)

, which lies on the circumference of the circle.

The Normal

The Radial Connection

A fundamental geometric truth is that the normal to a circle at any point always passes through its center. The normal line connects the center

A (2, 3)

and the point

P (5, 7)

The slope of the normal,

m_{N}

, is calculated as:

m_{N} = \frac{7 - 3}{5 - 2} = \frac{4}{3}

Using the point-slope form

y - y_{1} = m (x - x_{1})

, the equation of the normal is

y - 7 = \frac{4}{3} (x - 5)

. Simplifying this, we obtain:

4 x - 3 y + 1 = 0

To find the x-intercept

M

, we set

y = 0

, which yields

4 x + 1 = 0

, or

x = - \frac{1}{4}

. Thus, the vertex

M

(- \frac{1}{4}, 0)

The Tangent

The Perpendicular Dance

The tangent is perpendicular to the normal at the point of contact. Therefore, the slope of the tangent

m_{T}

is the negative reciprocal of

m_{N}

m_{T} = - \frac{3}{4}

Using the point

P (5, 7)

, the equation of the tangent is

y - 7 = - \frac{3}{4} (x - 5)

. Expanding and rearranging gives:

3 x + 4 y - 43 = 0

To find the x-intercept

N

, we set

y = 0

, resulting in

3 x = 43

, or

x = \frac{43}{3}

. Thus, the vertex

N

(\frac{43}{3}, 0)

The Final Area Calculation

We now consider the triangle

P M N

with vertices

P (5, 7)

M (- \frac{1}{4}, 0)

, and

N (\frac{43}{3}, 0)

. The base

b

lies on the x-axis:

b = \frac{43}{3} - (- \frac{1}{4}) = \frac{43}{3} + \frac{1}{4} = \frac{172 + 3}{12} = \frac{175}{12}

The height

h

of the triangle is the perpendicular distance from

P

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

P

, so

h = 7

. The area

A

is given by:

A = \frac{1}{2} \times base \times height = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}

The problem asks for the value of

24 A

. Multiplying our result by

24

, we find:

24A = 1225

Common Traps & FAQs

Why does the normal always pass through the center? Isn't that just a property of lines?

It is a fundamental property of circles! The tangent is perpendicular to the radius at the point of contact. Since the normal is perpendicular to the tangent, it must coincide with the radial line, which by definition passes through the center

(2, 3)

The question mentions the 'positive x-axis', but my intercept is negative. Did I do something wrong?

You are sharp! This is exactly why this question was controversial. The normal's x-intercept is indeed negative. In competitive exams, if a question is ambiguous, trust your derivation. The area calculation remains valid regardless of the sign of the intercept.

Can I use the determinant formula for the area of a triangle with vertices

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Absolutely! That is a powerful tool. You can use the determinant method: Area

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

. It will yield the same result as the base-height method, confirming your answer.

Animated Solution for Mathematics - Circles: If the area of the triangle formed by the positive x -axis, the normal and the tangent to the circle $(x - 2)^{2} + (y - 3)^{2} = 25$ at the point (5,7) is A, then 24 A is equal to Note: NTA has dropped this question in the final official answer key.

Enter Numerical Value:

Visualized Solution

Given Circle and Point $P (5, 7)$

Center and Radius

Slope of Normal $m_{N}$

Equation of Normal

Normal's X-intercept $M$

Slope of Tangent $m_{T}$

Equation of Tangent

Tangent's X-intercept $N$

Visualizing Triangle $P M N$

Length of Base $M N$

Height of Triangle $h$

Area of Triangle $A$

Final Value of $24 A$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal

The Tangent

The Final Area Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Given Circle and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(5,7)

Center and Radius

Slope of Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mN​

Equation of Normal

Normal's X-intercept .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Slope of Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mT​

Equation of Tangent

Tangent's X-intercept .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PMN

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

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

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

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 24A

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal

The Tangent

The Final Area Calculation

Similar Questions

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.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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the centre of a circle C be (α,β) and its radius r<8. Let 3x+4y=24 and 3x−4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α−β+r) is equal to

Given Circle and Point $P (5, 7)$

Slope of Normal $m_{N}$

Normal's X-intercept $M$

Slope of Tangent $m_{T}$

Tangent's X-intercept $N$

Visualizing Triangle $P M N$

Length of Base $M N$

Height of Triangle $h$

Area of Triangle $A$

Final Value of $24 A$

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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

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