JEE Main 2023 (06 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: A circle passing through the point $P (α, β)$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$ . The point $P$ is above the line $A B$ . The point $Q$ on the line segment $A B$ is the foot of perpendicular from $P$ on $A B$ . If $P Q$ is equal to 11 units, then the value of $α β$ is _______

Enter Numerical Value:

Progressive Hints

Hint 1

Visualize the circle in the first quadrant. Since it touches both the x-axis and y-axis, the distance from the center to both axes must be the radius

r

. This immediately fixes the center at

(r, r)

. Can you write the equation of this circle now?

Hint 2

You have two pieces of information: the point

P (α, β)

lies on the circle, and the perpendicular distance from

P

to the line

A B

is 11. Use the circle's equation to create a relationship between

α, β,

and

r

. Then, use the perpendicular distance formula for the line

x + y - r = 0

Hint 3

When you expand the squared perpendicular distance equation, you will get a term

(α + β - r)^{2}

. Compare this expansion with the equation you derived from the fact that

P

lies on the circle. You will find that the terms involving

r

cancel out perfectly, leaving you with the value of

α β

Visualized Solution

Circle Touching Coordinate Axes

Let the radius of the circle be $r$ .
Since it touches both axes in the first quadrant, its center is $(r, r)$ .

Equation of the Circle

Using the standard circle equation: $(x - h)^{2} + (y - k)^{2} = R^{2}$
Substitute center $(r, r)$ and radius $r$ :
$(x - r)^{2} + (y - r)^{2} = r^{2}$

Point $P$ on the Circle

The problem states that point $P (α, β)$ lies on the circle.
Substitute $x = α$ and $y = β$ :
$(α - r)^{2} + (β - r)^{2} = r^{2}$

Expanding the Circle Equation

Expand the squared terms:
$α^{2} - 2 r α + r^{2} + β^{2} - 2 r β + r^{2} = r^{2}$
Simplify to get Relation (1):
$α^{2} + β^{2} - 2 r α - 2 r β + r^{2} = 0 \dots (1)$

Points of Contact and Line $A B$

The circle touches the x-axis at $A (r, 0)$ and the y-axis at $B (0, r)$ .
Equation of line $A B$ using intercept form:
$\frac{x}{r} + \frac{y}{r} = 1 ⟹ x + y - r = 0$

Perpendicular Distance $P Q$

Point $Q$ is the foot of the perpendicular from $P (α, β)$ to line $A B$ .
The length of this perpendicular is given as $P Q = 11$ .

Applying the Distance Formula

The perpendicular distance from $(x_{1}, y_{1})$ to $A x + B y + C = 0$ is $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ .
Substitute $P (α, β)$ and line $x + y - r = 0$ :
$\frac{∣ α + β - r ∣}{1 ^{2} + 1 ^{2}} = 11$

Squaring the Distance Equation

Simplify the denominator: $1^{2} + 1^{2} = 2$
Equation becomes: $\frac{∣ α + β - r ∣}{2} = 11$
Square both sides to remove the absolute value:
$\frac{( α + β - r ) ^{2}}{2} = 121$

Expanding the Squared Term

Multiply by 2: $(α + β - r)^{2} = 242$
Expand using $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ :
$α^{2} + β^{2} + r^{2} + 2 α β - 2 r α - 2 r β = 242$

Rearranging to Use Relation (1)

Group terms to match Relation (1):
$(α^{2} + β^{2} - 2 r α - 2 r β + r^{2}) + 2 α β = 242$
Recall Relation (1): $α^{2} + β^{2} - 2 r α - 2 r β + r^{2} = 0$

Final Calculation

Substitute $0$ for the grouped terms:
$0 + 2 α β = 242$
Divide by 2:
$α β = 121$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

A circle nestled in the first quadrant that is tangent to both the

x

-axis and the

y

-axis possesses a center at

(r, r)

, where

r

is the radius of the circle.

The distance from the center to both axes is exactly

r

. Consequently, the equation of this circle is defined as:

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

The Point of Contact

Consider a point

P (α, β)

that lies on the circumference of this circle. By substituting these coordinates into the circle's equation, we obtain:

(α - r)^{2} + (β - r)^{2} = r^{2}

Expanding this expression yields:

α^{2} - 2 r α + r^{2} + β^{2} - 2 r β + r^{2} = r^{2}

By simplifying the terms, we arrive at a fundamental relationship, which we shall denote as Relation (1):

α^{2} + β^{2} - 2 r α - 2 r β + r^{2} = 0

The Line of Intersection

The circle touches the axes at points

A (r, 0)

and

B (0, r)

. The line segment

A B

connecting these points follows the intercept form equation:

\frac{x}{r} + \frac{y}{r} = 1 ⟹ x + y - r = 0

We are given that the perpendicular distance from point

P (α, β)

to this line is

11

. Applying the standard perpendicular distance formula:

\frac{∣ α + β - r ∣}{1 ^{2} + 1 ^{2}} = 11

The Algebraic Symphony

To resolve the absolute value and the radical, we square both sides of the equation:

\frac{( α + β - r ) ^{2}}{2} = 121 ⟹ (α + β - r)^{2} = 242

Expanding the trinomial

(α + β - r)^{2}

using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a

, we get:

α^{2} + β^{2} + r^{2} + 2 α β - 2 r α - 2 r β = 242

Observe that the terms

α^{2} + β^{2} - 2 r α - 2 r β + r^{2}

are exactly equal to

0

, as established in Relation (1). Substituting this into our expanded equation, the expression collapses to:

0 + 2 α β = 242

Dividing by

2

, we reach the final result:

α β = 121

Common Traps & FAQs

Wait, why is the center

(r, r)

? Can't it be

(r, - r)

That is a fantastic observation! In coordinate geometry, we must always respect the constraints. The problem explicitly states the circle is in the first quadrant. If the center were

(r, - r)

, the circle would dip into the fourth quadrant. Since it touches both axes in the first quadrant, the center must be equidistant from both axes, hence

(r, r)

is the only valid position.

Do I need to solve for

r

to find

α β

That is the beauty of this problem! You might feel the urge to find the specific value of

r

, but notice that

r

is an auxiliary variable. The algebraic structure of the problem is designed so that

r

cancels out. Don't hunt for

r

; hunt for the relationship between the variables. Trust the algebra!

Why is the perpendicular distance formula

\frac{∣ A x + B y + C ∣}{A ^{2} + B ^{2}}

This is a fundamental tool in coordinate geometry. It represents the shortest distance from a point to a line. The denominator

A^{2} + B^{2}

normalizes the equation of the line. Think of it as projecting the vector from the origin to the point onto the normal vector of the line. It is the bridge between geometry and algebra.

Enter Numerical Value:

Visualized Solution

Circle Touching Coordinate Axes

Equation of the Circle

Point $P$ on the Circle

Expanding the Circle Equation

Points of Contact and Line $A B$

Perpendicular Distance $P Q$

Applying the Distance Formula

Squaring the Distance Equation

Expanding the Squared Term

Rearranging to Use Relation (1)

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Point of Contact

The Line of Intersection

The Algebraic Symphony

Similar Questions

Let a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}, k > 0$ , touch the x-axis at $(1, 0)$ . If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is 2, then the value of $h + k + r$ is equal to ____.

If $P$ is a point on the circle $x^{2} + y^{2} = 4$ , $Q$ is a point on the straight line $5 x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $P Q$ , then 13 times the sum of abscissa of all such points $P$ is .........

Let $P Q$ be a diameter of the circle $x^{2} + y^{2} = 9$ . If $α$ and $β$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $α β$ is

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2} - 4 x - 6 = 0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2} + 2 y - 7 = 0$ . If $P Q$ is a diameter of the circle $x^{2} + y^{2} + 2 a x + 2 b y + c = 0$ , then the value of $(a + b - c)$ is

The abscissa of the two points $A$ and $B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and their ordinates are the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Find the equation and the radius of the circle with $A B$ as diameter.

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________

Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at $(- 1, 0)$ and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the $y$ -axis at the point $P$ , then the square of the distance of $P$ from the centre of $C_{1}$ is :

Enter Numerical Value:

Visualized Solution

Circle Touching Coordinate Axes

Equation of the Circle

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

Expanding the Circle Equation

Points of Contact and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

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

Applying the Distance Formula

Squaring the Distance Equation

Expanding the Squared Term

Rearranging to Use Relation (1)

Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Point of Contact

The Line of Intersection

The Algebraic Symphony

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the locus of the mid points of the chords of circle x2+(y−1)2=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of PQ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3​)x+2(4−6​)y=k+63​+86​,k>0, touch internally at the point P(α,β), then (α+3​)2+(β+6​)2 is equal to ________

Point $P$ on the Circle

Points of Contact and Line $A B$

Perpendicular Distance $P Q$

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + 2 (4 - 6) y = k + 63 + 86, k > 0$ , touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to ________