JEE Main 2022 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: A circle $C_{1}$ passes through the origin $O$ and has diameter 4 on the positive x-axis. The line $y = 2 x$ gives a chord $O A$ of a circle $C_{1}$ . Let $C_{2}$ be the circle with $O A$ as a diameter. If the tangent to $C_{2}$ at the point $A$ meets the x-axis at $P$ and y-axis at $Q$ , then $Q A : A P$ is equal to :

Q: Wait, why do we reject the negative root when solving for the intersection of the line and the circle?

Great observation! When we solved $5x^2 - 4x = 0$, we got $x=0$ and $x=4/5$. The $x=0$ solution corresponds to the origin $O$, which is where the line $y=2x$ starts. Since $A$ is the other intersection point (the chord endpoint), we naturally select the non-zero value, $x=4/5$.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry of circle

C_{1}

. It passes through the origin

O (0, 0)

and its diameter lies on the positive x-axis with length

4

. This means the center is at

(2, 0)

and the radius is

2

. Can you write the equation of

C_{1}

using this information?

Hint 2

Hint 2 (The Arsenal): Recall the property of a tangent to a circle at the endpoint of a diameter. If

O A

is the diameter of circle

C_{2}

, the tangent at

A

is perpendicular to the line segment

O A

. Use the slope of

O A

to find the slope of the tangent.

Hint 3

Hint 3 (The Execution): Instead of using the distance formula for

Q A

and

A P

, consider the x-coordinates of points

Q

A

, and

P

. Since they all lie on the same line, the ratio

Q A : A P

is equal to the ratio of the differences in their x-coordinates:

∣ x_{A} - x_{Q} ∣ : ∣ x_{P} - x_{A} ∣

Visualized Solution

Visualizing Circle $C_{1}$

Circle $C_{1}$ passes through origin $O (0, 0)$ .
Diameter is $4$ on the positive x-axis.

Equation of Circle $C_{1}$

Center of $C_{1}$ is $(2, 0)$ and radius $r = 2$ .
Equation: $(x - 2)^{2} + (y - 0)^{2} = 2^{2}$
$\Rightarrow x^{2} + y^{2} - 4 x = 0$

The Chord $O A$

Line $L : y = 2 x$ intersects $C_{1}$ .
This forms the chord $O A$ .

Finding Intersection Point $A$

Substitute $y = 2 x$ into $x^{2} + y^{2} - 4 x = 0$ .
$x^{2} + (2 x)^{2} - 4 x = 0$
$5 x^{2} - 4 x = 0$

Coordinates of Point $A$

$x (5 x - 4) = 0 \Rightarrow x = 0$ or $x = \frac{4}{5}$
For point $A$ , $x = \frac{4}{5}$ .
Substitute $x$ in $y = 2 x$ : $y = \frac{8}{5}$ .
Point $A = (\frac{4}{5}, \frac{8}{5})$ .

Introducing Circle $C_{2}$

Let $C_{2}$ be the circle with $O A$ as its diameter.

The Tangent at $A$

A tangent is drawn to $C_{2}$ at point $A$ .
Property: Tangent at the end of a diameter is perpendicular to the diameter.
Therefore, the tangent is perpendicular to $O A$ .

Slope of the Tangent

Slope of $O A$ ( $m_{1}$ ) $= 2$ (from $y = 2 x$ ).
Slope of tangent ( $m_{2}$ ) $= - \frac{1}{m _{1}} = - \frac{1}{2}$ .

Equation of the Tangent

Equation of tangent at $A (\frac{4}{5}, \frac{8}{5})$ with slope $- \frac{1}{2}$ :
$y - \frac{8}{5} = - \frac{1}{2} (x - \frac{4}{5})$
$2 y - \frac{16}{5} = - x + \frac{4}{5}$
$x + 2 y = \frac{20}{5} \Rightarrow x + 2 y = 4$

Finding Intercepts $P$ and $Q$

Tangent intersects x-axis at $P$ and y-axis at $Q$ .
For $P$ (x-axis), set $y = 0$ : $x = 4 \Rightarrow P (4, 0)$ .
For $Q$ (y-axis), set $x = 0$ : $2 y = 4 \Rightarrow y = 2 \Rightarrow Q (0, 2)$ .

The Smart Way to Find Ratio $Q A : A P$

We need the ratio $Q A : A P$ .
Using distance formula is lengthy.
Smart Trick: For points on a line, the ratio of segments equals the ratio of their x-coordinate differences.
Ratio $= ∣ x_{A} - x_{Q} ∣ : ∣ x_{P} - x_{A} ∣$

Calculating the Final Ratio

$x_{Q} = 0$ , $x_{A} = \frac{4}{5}$ , $x_{P} = 4$
$Q A : A P = ∣ x_{A} - x_{Q} ∣ : ∣ x_{P} - x_{A} ∣$
$Q A : A P = ∣ \frac{4}{5} - 0∣ : ∣4 - \frac{4}{5} ∣$
$Q A : A P = \frac{4}{5} : \frac{16}{5} = 1 : 4$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out at a circle

C_{1}

that touches the origin and stretches along the positive x-axis. The problem states the diameter is

4

, which anchors our circle.

With the diameter on the x-axis, the center must be at

(2, 0)

and the radius

r = 2

. The equation of this circle,

C_{1}

, is

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

, which simplifies to:

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

The Intersection

Finding Point

A

We introduce a line,

y = 2 x

, which cuts through our circle to create a chord

O A

. To find where this line meets the circle, we substitute

y = 2 x

into the circle equation:

x^{2} + (2 x)^{2} - 4 x = 0

This yields

5 x^{2} - 4 x = 0

. Solving this, we find

x = 0

(the origin) and

x = 4/5

Plugging

x = 4/5

back into

y = 2 x

, we get

y = 8/5

. Thus, point

A

is located at

(4/5, 8/5)

The Tangent and the Geometry of $C_{2}$

Next, we define a new circle,

C_{2}

, with

O A

as its diameter. We are interested in the tangent to this circle at point

A

The tangent at the endpoint of a diameter is always perpendicular to that diameter. Since the slope of

O A

2

, the slope of our tangent line must be the negative reciprocal,

- 1/2

Using the point-slope form with point

A (4/5, 8/5)

, the equation of the tangent becomes:

y - \frac{8}{5} = - \frac{1}{2} (x - \frac{4}{5})

Multiplying by

2

and rearranging, we get

2 y - 16/5 = - x + 4/5

, which simplifies to:

x + 2 y = 4

The Final Ratio

A Stroke of Elegance

The tangent line intersects the x-axis at

P

and the y-axis at

Q

. Setting

y = 0

gives

x = 4

, so

P (4, 0)

. Setting

x = 0

gives

2 y = 4

, so

y = 2

, meaning

Q (0, 2)

We need the ratio

Q A : A P

. Since

Q

A

, and

P

are collinear, the ratio of the segments

Q A : A P

is identical to the ratio of the differences in their x-coordinates:

∣ x_{A} - x_{Q} ∣ : ∣ x_{P} - x_{A} ∣

Substituting our values:

∣ \frac{4}{5} - 0∣ : ∣4 - \frac{4}{5} ∣ = \frac{4}{5} : \frac{16}{5}

The denominators cancel, leaving us with

4 : 16

, or

1 : 4

The final ratio is $1 : 4$ .

Common Traps & FAQs

Wait, why do we reject the negative root when solving for the intersection of the line and the circle?

Great observation! When we solved

5 x^{2} - 4 x = 0

, we got

x = 0

and

x = 4/5

. The

x = 0

solution corresponds to the origin

O

, which is where the line

y = 2 x

starts. Since

A

is the other intersection point (the chord endpoint), we naturally select the non-zero value,

x = 4/5

Why can't we just use the distance formula to find the ratio

Q A : A P

You absolutely can! However, in JEE Advanced, time is your most precious resource. Calculating the distance between two points involves square roots and potentially messy fractions. By using the property that the ratio of segments on a line is equal to the ratio of the differences of their x-coordinates (or y-coordinates), we bypass the radicals entirely. It is a much faster and more elegant approach.

How do we know the tangent at

A

is perpendicular to

O A

This is a fundamental geometric theorem: the tangent to a circle at any point is perpendicular to the radius at that point. Since

O A

is the diameter of circle

C_{2}

, the line segment

O A

acts as the radius (or diameter) connecting the center to the point of tangency

A

. Thus, the tangent must be perpendicular to

O A

Select Answer:

Visualized Solution

Visualizing Circle $C_{1}$

Equation of Circle $C_{1}$

The Chord $O A$

Finding Intersection Point $A$

Coordinates of Point $A$

Introducing Circle $C_{2}$

The Tangent at $A$

Slope of the Tangent

Equation of the Tangent

Finding Intercepts $P$ and $Q$

The Smart Way to Find Ratio $Q A : A P$

Calculating the Final Ratio

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Intersection

The Tangent and the Geometry of $C_{2}$

The Final Ratio

Similar Questions

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ 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

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 ________

List-I

List-II

Let the tangent to the circle $x^{2} + y^{2} = 25$ at the point R(3, 4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then $r^{2}$ is equal to :

Let $O$ be the origin and $O P$ and $O Q$ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $O P Q$ passes through the point $(α, \frac{1}{2})$ , then a value of $α$ is

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

Three circles touch one another externally. The tangent at their point of contact meet at a point whose distance from a point of contact is $4$ . Find the ratio of the product of the radii to the sum of the radii of the circles.

Select Answer:

Visualized Solution

Visualizing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Equation of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

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

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

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

Introducing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

The Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Slope of the Tangent

Equation of the Tangent

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

The Smart Way to Find Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QA:AP

Calculating the Final Ratio

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Intersection

The Tangent and the Geometry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

The Final Ratio

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the circle C1​:x2+y2−2x−1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C2​ whose centre is (3, -2). The radius of C2​ 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

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

Visualizing Circle $C_{1}$

Equation of Circle $C_{1}$

The Chord $O A$

Finding Intersection Point $A$

Coordinates of Point $A$

Introducing Circle $C_{2}$

The Tangent at $A$

Finding Intercepts $P$ and $Q$

The Smart Way to Find Ratio $Q A : A P$

The Tangent and the Geometry of $C_{2}$

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ 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

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to: