JEE Advanced 2019

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: A line $y = m x + 1$ intersects the circle $(x - 3)^{2} + (y + 2)^{2} = 25$ at the points $P$ and $Q$ . If the midpoint of the line segment $P Q$ has x-coordinate $- \frac{3}{5}$ , then which one of the following options is correct?

Q: Wait, why do we assume the midpoint $R$ lies on the line $y = mx + 1$?

That is a great observation! By definition, the chord $PQ$ is a segment of the line $y = mx + 1$. Since the midpoint $R$ of the chord must lie on the chord itself, it must satisfy the equation of the line. It is the bridge that connects the geometry of the circle to the algebra of the line.

Q: Why can't we just find the intersection points $P$ and $Q$ by substituting the line into the circle equation?

You absolutely could! You would get a quadratic in $x$, and you could use the sum of roots formula to find the midpoint. However, that path is algebraically heavy and prone to calculation errors. Using the perpendicularity property is the 'elegant' way—it bypasses the messy quadratic roots entirely.

Q: Is it always true that the line from the center to the midpoint is perpendicular to the chord?

Yes, always! This is a fundamental theorem in circle geometry. Think of the triangle $\triangle CPQ$. Since $CP = CQ$ (both are radii), the triangle is isosceles. The median from the vertex $C$ to the base $PQ$ is also the altitude. That is why the perpendicularity holds!

Q: What if the line was vertical? Would the slope product rule still work?

Excellent catch. If the line were vertical, its slope would be undefined. In that case, the perpendicular line (the radius) would be horizontal (slope $0$). The product rule $m_1 \cdot m_2 = -1$ is a specific tool for non-vertical lines, so always check your line's orientation first!

Q: How do I know which option to pick if I get two values for $m$?

The question asks which interval contains the possible values of $m$. Since we found $m=2$ and $m=3$, we look for the interval that encompasses these values. Always check if your calculated values satisfy the constraints of the options provided.

Select Answer:

Visualized Solution

$T h e C i r c l e an d i t s C e n t er$

Circle equation: $(x - 3)^{2} + (y + 2)^{2} = 25$
Center $C (3, - 2)$
Radius $r = 5$

$T h e I n t er sec t in g L in e$

Line equation: $y = m x + 1$
Intersects the circle at points $P$ and $Q$ .

$M i d p o in t o f t h e C h or d$

Let $R$ be the midpoint of chord $P Q$ .
Given x-coordinate of $R$ : $x_{R} = - \frac{3}{5}$

$F in d in g t h ey - coor d ina t eo f R$

Point $R$ lies on the line $y = m x + 1$ .
Substitute $x = - \frac{3}{5}$ :
$y_{R} = m (- \frac{3}{5}) + 1$

$C oor d ina t eso f M i d p o in tR$

$y_{R} = 1 - \frac{3 m}{5}$
Midpoint $R \equiv (- \frac{3}{5}, 1 - \frac{3 m}{5})$

$T h e P er p e n d i c u l a r i t y P r o p er t y$

Geometric Property: The line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.
Therefore, $C R ⊥ P Q$ .

$S l o p eso f P er p e n d i c u l a r L in es$

Since $C R ⊥ P Q$ , their slopes multiply to $- 1$ .
$(Slope of C R) \times (Slope of P Q) = - 1$
We know the slope of $P Q$ is $m$ .

$S e tt in g u pS l o p eo f C R$

$C (3, - 2)$ and $R (- \frac{3}{5}, 1 - \frac{3 m}{5})$
$Slope of C R = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
$Slope of C R = \frac{( 1 - \frac{3 m}{5} ) - ( - 2 )}{- \frac{3}{5} - 3}$

$S im pl i f y in g t h e S l o p eo f C R$

Numerator: $1 - \frac{3 m}{5} + 2 = 3 - \frac{3 m}{5} = \frac{15 - 3 m}{5}$
Denominator: $- \frac{3}{5} - 3 = \frac{- 3 - 15}{5} = - \frac{18}{5}$
$Slope of C R = \frac{\frac{15 - 3 m}{5}}{- \frac{18}{5}}$

$F ina l S l o p eo f C R$

Cancel the common denominator $5$ .
$Slope of C R = \frac{15 - 3 m}{- 18}$
Divide by $- 3$ :
$Slope of C R = \frac{m - 5}{6}$

$A ppl y in g t h e P er p e n d i c u l a r C o n d i t i o n$

$(Slope of C R) \times (Slope of P Q) = - 1$
$(\frac{m - 5}{6}) \cdot m = - 1$

$F or min g t h e Q u a d r a t i c E q u a t i o n$

Multiply both sides by $6$ :
$m (m - 5) = - 6$
Expand and rearrange:
$m^{2} - 5 m + 6 = 0$

$S o l v in g f or m$

Factorize the quadratic equation:
$(m - 2) (m - 3) = 0$
The possible values for slope are:
$m = 2$ or $m = 3$

$C o n c l u s i o nan d F ina l A n s w er$

We found $m = 2$ or $m = 3$ .
Checking the given options:
Both values lie in the interval $2 \leq m < 4$ .
Correct Option: $[2, 4)$

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We are given a circle defined by the equation

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

. From this, we identify the center

C

(3, - 2)

and the radius

r

5

A line

y = m x + 1

intersects this circle to form a chord

P Q

. We are given that the midpoint

R

of this chord has an

x

-coordinate of

x_{R} = - \frac{3}{5}

The Bridge Between Algebra and Geometry

The midpoint

R

must lie on the line

y = m x + 1

. By substituting

x_{R} = - \frac{3}{5}

into the line equation, we determine the

y

-coordinate of

R

y_{R} = m (- \frac{3}{5}) + 1 = 1 - \frac{3 m}{5}

Thus, the coordinates of the midpoint

R

are

(- \frac{3}{5}, 1 - \frac{3 m}{5})

. This point serves as our primary anchor for the geometric derivation.

The Power of Perpendicularity

A fundamental property of circles is that the line segment connecting the center

C

to the midpoint

R

of a chord is always perpendicular to the chord itself. Therefore, the product of the slope of

C R

and the slope of the chord

P Q

must be

- 1

We calculate the slope of

C R

using the coordinates

C (3, - 2)

and

R (- \frac{3}{5}, 1 - \frac{3 m}{5})

Slope of C R = \frac{y _{R} - y _{C}}{x _{R} - x _{C}} = \frac{( 1 - \frac{3 m}{5} ) - ( - 2 )}{- \frac{3}{5} - 3}

Simplifying the expression above:

Slope of C R = \frac{3 - \frac{3 m}{5}}{- \frac{18}{5}} = \frac{\frac{15 - 3 m}{5}}{- \frac{18}{5}} = \frac{15 - 3 m}{- 18} = \frac{m - 5}{6}

The Final Convergence

Applying the perpendicularity condition

(Slope of C R) \times (Slope of P Q) = - 1

, and knowing the slope of the line

P Q

m

, we obtain:

(\frac{m - 5}{6}) \cdot m = - 1

Multiplying both sides by

6

yields the quadratic equation:

m^{2} - 5 m + 6 = 0

Factoring the quadratic gives

(m - 2) (m - 3) = 0

. Consequently, the possible values for the slope are $m = 2$ or $m = 3$ .

Common Traps & FAQs

Wait, why do we assume the midpoint

R

lies on the line

y = m x + 1

That is a great observation! By definition, the chord

P Q

is a segment of the line

y = m x + 1

. Since the midpoint

R

of the chord must lie on the chord itself, it must satisfy the equation of the line. It is the bridge that connects the geometry of the circle to the algebra of the line.

Why can't we just find the intersection points

P

and

Q

by substituting the line into the circle equation?

You absolutely could! You would get a quadratic in

x

, and you could use the sum of roots formula to find the midpoint. However, that path is algebraically heavy and prone to calculation errors. Using the perpendicularity property is the 'elegant' way—it bypasses the messy quadratic roots entirely.

Is it always true that the line from the center to the midpoint is perpendicular to the chord?

Yes, always! This is a fundamental theorem in circle geometry. Think of the triangle

△ C P Q

. Since

C P = C Q

(both are radii), the triangle is isosceles. The median from the vertex

C

to the base

P Q

is also the altitude. That is why the perpendicularity holds!

What if the line was vertical? Would the slope product rule still work?

Excellent catch. If the line were vertical, its slope would be undefined. In that case, the perpendicular line (the radius) would be horizontal (slope

0

). The product rule

m_{1} \cdot m_{2} = - 1

is a specific tool for non-vertical lines, so always check your line's orientation first!

How do I know which option to pick if I get two values for

m

The question asks which interval contains the possible values of

m

. Since we found

m = 2

and

m = 3

, we look for the interval that encompasses these values. Always check if your calculated values satisfy the constraints of the options provided.

Animated Solution for Mathematics - Circles: A line $y = m x + 1$ intersects the circle $(x - 3)^{2} + (y + 2)^{2} = 25$ at the points $P$ and $Q$ . If the midpoint of the line segment $P Q$ has x-coordinate $- \frac{3}{5}$ , then which one of the following options is correct?

Select Answer:

Visualized Solution

$T h e C i r c l e an d i t s C e n t er$

$T h e I n t er sec t in g L in e$

$M i d p o in t o f t h e C h or d$

$F in d in g t h ey - coor d ina t eo f R$

$C oor d ina t eso f M i d p o in tR$

$T h e P er p e n d i c u l a r i t y P r o p er t y$

$S l o p eso f P er p e n d i c u l a r L in es$

$S e tt in g u pS l o p eo f C R$

$S im pl i f y in g t h e S l o p eo f C R$

$F ina l S l o p eo f C R$

$A ppl y in g t h e P er p e n d i c u l a r C o n d i t i o n$

$F or min g t h e Q u a d r a t i c E q u a t i o n$

$S o l v in g f or m$

$C o n c l u s i o nan d F ina l A n s w er$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Bridge Between Algebra and Geometry

The Power of Perpendicularity

The Final Convergence

Similar Questions

Let $C$ be a circle with radius $10$ units and centre at the origin. Let the line $x + y = 2$ intersects the circle $C$ at the points $P$ and $Q$ . Let $M N$ be a chord of $C$ of length 2 unit and slope -1. Then, a distance (in units) between the chord $P Q$ and the chord $M N$ is

If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of the radius $r$ , then $r$ is equal to :

The points of intersection of the line $4 x - 3 y - 10 = 0$ and the circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ are ......... and .........

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Let a circle $C$ touch the lines $L_{1} : 4 x - 3 y + K_{1} = 0$ and $L_{2} : 4 x - 3 y + K_{2} = 0, K_{1}, K_{2} \in R$ . If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(- 1, 2)$ and $L_{2}$ at $(3, - 6)$ , then the equation of the circle $C$ 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 ________

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to

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 ____.

Select Answer:

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingthey−coordinateofR

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Bridge Between Algebra and Geometry

The Power of Perpendicularity

The Final Convergence

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the locus of the mid-point of the line segment from the point (3,2) to a point on the circle, x2+y2=1 is a circle of the radius r, then r is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The points of intersection of the line 4x−3y−10=0 and the circle x2+y2−2x+4y−20=0 are ......... and .........

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle passes through the points (2,3) and (4,5). If its centre lies on the line, y−4x+3=0, then its radius is equal to

$T h e C i r c l e an d i t s C e n t er$

$T h e I n t er sec t in g L in e$

$M i d p o in t o f t h e C h or d$

$F in d in g t h ey - coor d ina t eo f R$

$C oor d ina t eso f M i d p o in tR$

$T h e P er p e n d i c u l a r i t y P r o p er t y$

$S l o p eso f P er p e n d i c u l a r L in es$

$S e tt in g u pS l o p eo f C R$

$S im pl i f y in g t h e S l o p eo f C R$

$F ina l S l o p eo f C R$

$A ppl y in g t h e P er p e n d i c u l a r C o n d i t i o n$

$F or min g t h e Q u a d r a t i c E q u a t i o n$

$S o l v in g f or m$

$C o n c l u s i o nan d F ina l A n s w er$

If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of the radius $r$ , then $r$ is equal to :

The points of intersection of the line $4 x - 3 y - 10 = 0$ and the circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ are ......... and .........

The circle passing through $(1, - 2)$ and touching the axis of $x$ at $(3, 0)$ also passes through the point

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 ________

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to