Wait, why do we assume $\angle PQR = 90^\circ$? Is that always true for any circle?

That is a fantastic question! Yes, it is always true. This is known as Thales' Theorem. It states that if $A, B,$ and $C$ are distinct points on a circle where the line $AC$ is a diameter, then the angle $\angle ABC$ is a right angle. It is a fundamental property of circles that turns a geometry problem into a simple slope calculation. Never forget this tool; it is a lifesaver in JEE exams!

JEE Main 2023 (25 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: Points $P (- 3, 2), Q (9, 10)$ and $R (α, 4)$ lie on a circle $C$ with $P R$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$ . If $S$ lies on the line $2 x - k y = 1$ , then $k$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Before you dive into the algebra, look at the geometry. You are given a diameter

P R

. Recall Thales' Theorem: any angle subtended by a diameter at the circumference is a right angle. This means

∠ P QR = 9 0^{\circ}

. This is your golden ticket to finding

α

Hint 2

Hint 2: The Arsenal. Since

P Q

and

QR

are perpendicular, the product of their slopes must be

- 1

. Use the slope formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

for both segments. Set their product to

- 1

and solve for the unknown

α

Hint 3

Hint 3: The Execution. Once you have the circle's center and radius, you need the tangents at

Q

and

R

. Do not derive them from scratch! Use the standard equation of a tangent at point

(x_{1}, y_{1})

for a circle

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, which is

x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0

. This is the fastest, most elegant path to the intersection point

S

Visualized Solution

Visualizing the Geometry

Points $P (- 3, 2)$ , $Q (9, 10)$ , and $R (α, 4)$ lie on circle $C$ .
$P R$ is the diameter of the circle.

The $9 0^{\circ}$ Property

By Thales' Theorem, the angle subtended by a diameter at any point on the circle is $9 0^{\circ}$ .
Therefore, $∠ P QR = 9 0^{\circ}$ .
This implies the line segments $P Q$ and $QR$ are perpendicular.

Slopes Setup

For perpendicular lines, the product of their slopes is $- 1$ .
$m_{P Q} \cdot m_{QR} = - 1$

Calculating Slope of $P Q$

Slope formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
$m_{P Q} = \frac{10 - 2}{9 - ( - 3 )}$
$m_{P Q} = \frac{8}{12} = \frac{2}{3}$

Calculating Slope of $QR$ and solving for $α$

$m_{QR} = \frac{4 - 10}{α - 9} = \frac{- 6}{α - 9}$
$(\frac{2}{3}) \cdot (\frac{- 6}{α - 9}) = - 1$
$\frac{- 4}{α - 9} = - 1 ⟹ α - 9 = 4$
$α = 13$ . Point $R$ is $(13, 4)$ .

Circle's Center and Radius

Center $M$ is the midpoint of diameter $P R$ .
$M = (\frac{- 3 + 13}{2}, \frac{2 + 4}{2}) = (5, 3)$
Radius squared $r^{2} = (13 - 5)^{2} + (4 - 3)^{2} = 8^{2} + 1^{2} = 65$

Equation of the Circle

Standard equation: $(x - h)^{2} + (y - k)^{2} = r^{2}$
$(x - 5)^{2} + (y - 3)^{2} = 65$
Expanding: $x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = 65$
$x^{2} + y^{2} - 10 x - 6 y - 31 = 0$

Tangent at $Q (9, 10)$

Equation of tangent at $(x_{1}, y_{1})$ is $T = 0$ .
$x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$
At $Q (9, 10)$ : $9 x + 10 y - 5 (x + 9) - 3 (y + 10) - 31 = 0$
$4 x + 7 y = 106$

Tangent at $R (13, 4)$

Applying $T = 0$ at $R (13, 4)$ .
$13 x + 4 y - 5 (x + 13) - 3 (y + 4) - 31 = 0$
$13 x - 5 x + 4 y - 3 y - 65 - 12 - 31 = 0$
$8 x + y = 108$

Finding Intersection Point $S$

Solve $4 x + 7 y = 106$ and $8 x + y = 108$ .
From the second equation: $y = 108 - 8 x$ .
Substitute into the first: $4 x + 7 (108 - 8 x) = 106$
$4 x + 756 - 56 x = 106 ⟹ - 52 x = - 650 ⟹ x = \frac{25}{2}$
$y = 108 - 8 (\frac{25}{2}) = 108 - 100 = 8$ .
Point $S$ is $(\frac{25}{2}, 8)$ .

The Final Condition

Point $S (\frac{25}{2}, 8)$ lies on the line $2 x - k y = 1$ .
Substitute $x = \frac{25}{2}$ and $y = 8$ into the line equation.
$2 (\frac{25}{2}) - k (8) = 1$

Solving for $k$

$25 - 8 k = 1$
$8 k = 24$
$k = 3$
Final Answer: $k = 3$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are given three points:

P (- 3, 2)

Q (9, 10)

, and

R (α, 4)

. These points lie on a circle

C

where

P R

is the diameter.

Our objective is to find the intersection point

S

of the tangents at

Q

and

R

, and subsequently determine the value of

k

given that

S

lies on the line

2 x - k y = 1

Phase 1

The Silent Guardian

Since

P R

is the diameter, by Thales' Theorem, the angle subtended by the diameter at any point on the circumference is

9 0^{\circ}

. Therefore,

∠ P QR = 9 0^{\circ}

, implying that the line segments

P Q

and

QR

are perpendicular.

The slope of

P Q

is calculated as:

m_{P Q} = \frac{10 - 2}{9 - ( - 3 )} = \frac{8}{12} = \frac{2}{3}

The slope of

QR

is:

m_{QR} = \frac{4 - 10}{α - 9} = \frac{- 6}{α - 9}

Since the product of the slopes of perpendicular lines is

- 1

, we have:

(\frac{2}{3}) \cdot (\frac{- 6}{α - 9}) = - 1

Simplifying this expression, we obtain

\frac{- 4}{α - 9} = - 1

, which leads to

α - 9 = 4

. Thus, $α = 13$ , and point

R

(13, 4)

Phase 2

Defining the Circle

The center

M

of the circle is the midpoint of the diameter

P R

M = (\frac{- 3 + 13}{2}, \frac{2 + 4}{2}) = (5, 3)

The radius squared,

r^{2}

, is the squared distance from the center

M (5, 3)

to point

R (13, 4)

r^{2} = (13 - 5)^{2} + (4 - 3)^{2} = 8^{2} + 1^{2} = 65

The equation of the circle is

(x - 5)^{2} + (y - 3)^{2} = 65

. Expanding this, we get:

x^{2} + y^{2} - 10 x - 6 y - 31 = 0

Phase 3

The Tangent Tango

To find the tangents at

Q (9, 10)

and

R (13, 4)

, we use the

T = 0

formula:

x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0

For point

Q (9, 10)

9 x + 10 y - 5 (x + 9) - 3 (y + 10) - 31 = 0

4 x + 7 y = 106

For point

R (13, 4)

13 x + 4 y - 5 (x + 13) - 3 (y + 4) - 31 = 0

8 x + y = 108

Phase 4

The Final Convergence

We solve the system of linear equations: 1)

4 x + 7 y = 106

8 x + y = 108

From the second equation,

y = 108 - 8 x

. Substituting this into the first equation:

4 x + 7 (108 - 8 x) = 106

4 x + 756 - 56 x = 106

- 52 x = - 650 \Rightarrow x = \frac{25}{2}

Substituting

x = 12.5

back into

y = 108 - 8 x

y = 108 - 8 (12.5) = 108 - 100 = 8

The intersection point

S

(\frac{25}{2}, 8)

. Since

S

lies on

2 x - k y = 1

2 (\frac{25}{2}) - k (8) = 1

25 - 8 k = 1

8 k = 24

The final value is $k = 3$ .

Common Traps & FAQs

Wait, why do we assume

∠ P QR = 9 0^{\circ}

? Is that always true for any circle?

That is a fantastic question! Yes, it is always true. This is known as Thales' Theorem. It states that if

A, B,

and

C

are distinct points on a circle where the line

A C

is a diameter, then the angle

∠ A B C

is a right angle. It is a fundamental property of circles that turns a geometry problem into a simple slope calculation. Never forget this tool; it is a lifesaver in JEE exams!

Can I just use the slope-intercept form

y = m x + c

to find the tangents? Why use the

T = 0

method?

You absolutely could, but it is a trap! Finding the slope of the tangent, then the equation, then solving for the intercept is a long, error-prone process. The

T = 0

method (the chord of contact or tangent equation) is a powerful, direct algebraic shortcut. It saves time and minimizes the chance of a calculation error. In competitive exams, efficiency is just as important as accuracy.

Why do we reject the negative root or worry about the sign of

α

In this specific problem, the algebra leads us to a linear equation for

α

, so we do not have to worry about rejecting roots. However, always keep an eye on the geometry. If you ever get two values for a coordinate, check if they satisfy the physical constraints of the problem (like the circle's orientation). Here, the math is straightforward, so trust your derivation!

Animated Solution for Mathematics - Circles: Points $P (- 3, 2), Q (9, 10)$ and $R (α, 4)$ lie on a circle $C$ with $P R$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$ . If $S$ lies on the line $2 x - k y = 1$ , then $k$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

The $9 0^{\circ}$ Property

Slopes Setup

Calculating Slope of $P Q$

Calculating Slope of $QR$ and solving for $α$

Circle's Center and Radius

Equation of the Circle

Tangent at $Q (9, 10)$

Tangent at $R (13, 4)$

Finding Intersection Point $S$

The Final Condition

Solving for $k$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 90∘ Property

Slopes Setup

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

Calculating Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR and solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Circle's Center and Radius

Equation of the Circle

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(9,10)

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R(13,4)

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Final Condition

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

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=x+2,4y=3x+6 and 3y=4x+1 be three tangent lines to the circle (x−h)2+(y−k)2=r2. Then h+k is equal to :

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