JEE Main 2018 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: 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

Q: Is there a faster way to find the center?

Yes, you could find the perpendicular bisector of chord $AB$ and find its intersection with the given line. Both methods are elegant, but the algebraic approach we used is robust and less prone to geometric visualization errors.

Q: What if the center coordinates were not integers?

That is a valid concern. If $h$ turned out to be a fraction, the radius calculation would still follow the same logic. Always trust the algebra; it is the language of geometry and will guide you to the correct answer regardless of the numbers.

Select Answer:

Visualized Solution

Visualizing the Given Points

Given points on the circle: $A (2, 3)$ and $B (4, 5)$
Objective: Find the radius $r$ of the circle.

The Center's Path

The center of the circle lies on the line: $y - 4 x + 3 = 0$
Rearranging for $y$ : $y = 4 x - 3$

Defining the Center $C (h, k)$

Let the center be $C (h, k)$ .
Since $C$ lies on $y = 4 x - 3$ , substitute $x = h$ and $y = k$ .
We get $k = 4 h - 3$ .
Thus, the center is $C (h, 4 h - 3)$ .

Applying the Radius Property

Property of a circle: Distance from the center to any point on the circumference is constant.
Therefore, $C A = C B = r$
Squaring both sides to remove square roots: $C A^{2} = C B^{2}$

Setting up the Distance Equation

Using the distance formula: $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$C A^{2} = (h - 2)^{2} + (4 h - 3 - 3)^{2}$
$C B^{2} = (h - 4)^{2} + (4 h - 3 - 5)^{2}$
Equating them: $(h - 2)^{2} + (4 h - 6)^{2} = (h - 4)^{2} + (4 h - 8)^{2}$

Expanding the Squares

LHS: $(h^{2} - 4 h + 4) + (16 h^{2} - 48 h + 36)$
RHS: $(h^{2} - 8 h + 16) + (16 h^{2} - 64 h + 64)$

Simplifying the Equation

Combine terms on LHS: $17 h^{2} - 52 h + 40$
Combine terms on RHS: $17 h^{2} - 72 h + 80$
The $17 h^{2}$ terms cancel out on both sides.
Remaining equation: $- 52 h + 40 = - 72 h + 80$

Solving for $h$

Rearranging terms: $72 h - 52 h = 80 - 40$
$20 h = 40$
$h = 2$

Finding the $y$ -coordinate $k$

Substitute $h = 2$ into $k = 4 h - 3$
$k = 4 (2) - 3$
$k = 8 - 3 = 5$
The exact center is $C (2, 5)$

Calculating the Radius $r$

Radius $r$ is the distance from $C (2, 5)$ to $A (2, 3)$ .
$r = (2 - 2)^{2} + (5 - 3)^{2}$
$r = 0^{2} + 2^{2}$
$r = 4 = 2$

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We are tasked with finding the properties of a circle passing through points

A (2, 3)

and

B (4, 5)

, with its center

C (h, k)

constrained to the line

y - 4 x + 3 = 0

Since the center

C (h, k)

lies on the line

y = 4 x - 3

, we can express the coordinates of the center as

C (h, 4 h - 3)

. This substitution effectively reduces our variables from two to one.

The Algebraic Dance

The fundamental property of a circle is that the distance from the center to any point on the circumference is equal to the radius

r

. Therefore, the distance

C A

must equal the distance

C B

, which implies

C A^{2} = C B^{2}

Applying the distance formula, we set up the following equation:

(h - 2)^{2} + (4 h - 3 - 3)^{2} = (h - 4)^{2} + (4 h - 3 - 5)^{2}

Simplifying the terms inside the parentheses, we obtain:

(h - 2)^{2} + (4 h - 6)^{2} = (h - 4)^{2} + (4 h - 8)^{2}

Expanding both sides of the equation:

(h^{2} - 4 h + 4) + (16 h^{2} - 48 h + 36) = (h^{2} - 8 h + 16) + (16 h^{2} - 64 h + 64)

Combining like terms, the quadratic

17 h^{2}

terms cancel out from both sides, leaving us with a linear equation:

- 52 h + 40 = - 72 h + 80

Solving for

h

20 h = 40 \Rightarrow h = 2

Substituting

h = 2

back into our expression for

k

k = 4 (2) - 3 = 5

Thus, the center of the circle is $C (2, 5)$ .

Final Calculation

The radius

r

is the distance from the center

C (2, 5)

to point

A (2, 3)

r = (2 - 2)^{2} + (5 - 3)^{2}

r = 0^{2} + 2^{2} = 2

The circle is defined by the center $(2, 5)$ and the radius $r = 2$ .

Common Traps & FAQs

Why do we equate

C A^{2}

and

C B^{2}

instead of just

C A

and

C B

Great question! We square both sides to eliminate the square root, which makes the algebra much cleaner and prevents us from dealing with messy radical expressions. It is a standard technique in coordinate geometry to simplify distance-based problems.

Is there a faster way to find the center?

Yes, you could find the perpendicular bisector of chord

A B

and find its intersection with the given line. Both methods are elegant, but the algebraic approach we used is robust and less prone to geometric visualization errors.

What if the center coordinates were not integers?

That is a valid concern. If

h

turned out to be a fraction, the radius calculation would still follow the same logic. Always trust the algebra; it is the language of geometry and will guide you to the correct answer regardless of the numbers.

Animated Solution for Mathematics - Circles: 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

Select Answer:

Visualized Solution

Visualizing the Given Points

The Center's Path

Defining the Center $C (h, k)$

Applying the Radius Property

Setting up the Distance Equation

Expanding the Squares

Simplifying the Equation

Solving for $h$

Finding the $y$ -coordinate $k$

Calculating the Radius $r$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Dance

Final Calculation

Similar Questions

Let $C$ be a circle passing through the points $A (2, - 1)$ and $B (3, 4)$ . The line segment $A B$ is not a diameter of $C$ . If $r$ is the radius of $C$ and its centre lies on the circle $(x - 5)^{2} + (y - 1)^{2} = \frac{13}{2}$ , then $r^{2}$ is equal to :

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 one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is

Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

If one of the diameters of the circle, given by the equation, $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ , is a chord of a circle $S$ , whose centre is at $(- 3, 2)$ , then the radius of $S$ is:

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle is

Let a circle C pass through the points $(4, 2)$ and (0, 2), and its centre lie on $y + 2 = 0$ Then the length of the chord, of the circle C, whose mid-point is $(1, 2)$ , 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 :

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

The diameter of the circle, whose centre lies on the line $x + y = 2$ in the first quadrant and which touches both the lines $x = 3$ and $y = 2$ , is

Animated Solution for Mathematics - Circles: .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

Select Answer:

Visualized Solution

Visualizing the Given Points

The Center's Path

Defining the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(h,k)

Applying the Radius Property

Setting up the Distance Equation

Expanding the Squares

Simplifying the Equation

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

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

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

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters of the circle x2+y2−2x−6y+6=0 is a chord to the circle with centre (2,1), then the radius of the circle is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equations of the circle passing through (−4,3) and touching the lines x+y=2 and x−y=2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3,2), then the radius of S is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x+3y+1=0 and 3x−y−4=0 lie along diameter of a circle of circumference 10π, then the equation of the circle is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a circle C pass through the points (4,2) and (0, 2), and its centre lie on y+2=0 Then the length of the chord, of the circle C, whose mid-point is (1,2), is:

.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; } If one of the diameters o of the circle x2+y2−2x−6y+6=0 is a chord of another circle 'C' whose center is at (2, 1), then its radius is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The diameter of the circle, whose centre lies on the line x+y=2 in the first quadrant and which touches both the lines x=3 and y=2, is

Animated Solution for Mathematics - Circles: 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

Defining the Center $C (h, k)$

Solving for $h$

Finding the $y$ -coordinate $k$

Calculating the Radius $r$

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is

Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

If one of the diameters of the circle, given by the equation, $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ , is a chord of a circle $S$ , whose centre is at $(- 3, 2)$ , then the radius of $S$ is:

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle is

Let a circle C pass through the points $(4, 2)$ and (0, 2), and its centre lie on $y + 2 = 0$ Then the length of the chord, of the circle C, whose mid-point is $(1, 2)$ , 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 :

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

The diameter of the circle, whose centre lies on the line $x + y = 2$ in the first quadrant and which touches both the lines $x = 3$ and $y = 2$ , is