If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

If the lines 2x - 3y = 5 and 3x - 4y = 7 are the d – Class 11 Mathematics

NCERT Exemplar · Q23

2x - 3y = 5

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

(x - h)^2 + (y - k)^2 = r^2

That's it. That's the standard form of the equation of a circle.

(x - h)^2 + (y - k)^2 = r^2

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …

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The center lies at the intersection of the two diameter lines, and the radius comes from the given area. The circle equation is $(x - 1)^2 + (y + 1)^2 = 49$ .

Why this approach works

A diameter of a circle is any chord that passes through the center. If two lines are both diameters, their point of intersection must be the center of the circle. Once we have the center and can find the radius from the area, the standard form $(x - h)^2 + (y - k)^2 = r^2$ gives us the equation immediately.

The area formula $A = \pi r^2$ connects the given area to the radius we need.

Solution

1. Find the center by solving the system of diameter equations

The center $(h, k)$ satisfies both diameter equations simultaneously:

2x - 3y = 5 \quad \text{...(i)}

3x - 4y = 7 \quad \text{...(ii)}

Multiply equation (i) by 3 and equation (ii) by 2:

6x - 9y = 15

6x - 8y = 14

Subtract the second from the first:

-9y + 8y = 15 - 14

-y = 1

y = -1

Substitute $y = -1$ into equation (i):

2x - 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

The center is at $(1, -1)$ .

2. Find the radius from the given area

The area of the circle is 154 square units:

\pi r^2 = 154

Using $\pi = \frac{22}{7}$ : …

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Q.If the lines 2x−3y=52x - 3y = 52x−3y=5 and 3x−4y=73x - 4y = 73x−4y=7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Why this approach works

Solution

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Q.If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

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