The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units. Then the equa

Q: Wait, why do we solve the intersection of the two lines? Is that always the center?

Yes, absolutely! By definition, a diameter is any chord that passes through the center of a circle. Since both lines are diameters, they must both pass through the center. Therefore, their unique point of intersection is the center of the circle. It is a beautiful, direct geometric property.

Q: Why do we use $\frac{22}{7}$ for $\pi$? Can't we just use $3.14$?

In JEE problems, you will often find that using $\frac{22}{7}$ leads to clean, integer results because the problem setters design the numbers to cancel out. Using $3.14$ might lead to messy decimals that make it harder to see the final answer. Always look for the most convenient form of constants!

Q: I keep getting the signs wrong in $(x - h)^2 + (y - k)^2 = r^2$. How do I avoid this?

This is a classic trap! Always write the center as $(h, k)$ first. If your center is $(1, -1)$, then $h=1$ and $k=-1$. When you plug them in, you get $(x - 1)^2 + (y - (-1))^2$. The double negative $(y - (-1))$ becomes $(y + 1)$. Take it one step at a time, and you will never miss a sign.

Question

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle of area 154 sq. units. Then the equation of this circle is

Options:

(A) $x^2 + y^2 + 2x - 2y = 62$

(B) $x^2 + y^2 + 2x - 2y = 47$

(C) $x^2 + y^2 - 2x + 2y = 47$

(D) $x^2 + y^2 - 2x + 2y = 62$

Sigma Percentile · Accepted Answer

### The Geometry of Circles: A Journey to the Center

Welcome, future engineer! Today, we are going to unravel the mystery of a circle defined by its diameters. Coordinate geometry is not just about plugging numbers into formulas; it is about visualizing the skeleton of a shape and bringing it to life.

Imagine you are standing on a coordinate plane, and you see two lines crossing. These aren't just any lines; they are the diameters of a circle. Let's find the heart of this circle.

### Phase 1: Finding the Center

We are given two lines: $2x - 3y = 5$ and $3x - 4y = 7$. As we discussed, these lines are diameters.

The geometric reality is that the center of the circle, let's call it $C(h, k)$, must lie on both lines. Therefore, the intersection point of these two lines is the center of our circle.

To find this, we solve the system of linear equations using the elimination method. Let's multiply the first equation by $3$ and the second by $2$ to align our $x$ terms:

$$3 	imes (2x - 3y = 5) \Rightarrow 6x - 9y = 15$$

$$2 	imes (3x - 4y = 7) \Rightarrow 6x - 8y = 14$$

Now, we subtract the second equation from the first. The $6x$ terms cancel out beautifully, leaving us with $-9y - (-8y) = 15 - 14$, which simplifies to $-y = 1$, or $y = -1$.

Substituting $y = -1$ back into our first equation, $2x - 3(-1) = 5$, we get $2x + 3 = 5$, which means $2x = 2$, so $x = 1$. We have found our center: $C(1, -1)$.

### Phase 2: Unlocking the Radius

Now that we have the center, we need the radius. The problem tells us the area is $154$ sq. units.

We know the area of a circle is $A = \pi r^2$. Setting this up, we have:

$$\frac{22}{7} r^2 = 154$$

By multiplying both sides by $\frac{7}{22}$, we get $r^2 = 154 	imes \frac{7}{22}$. Since $154$ divided by $22$ is $7$, we find $r^2 = 7 	imes 7 = 49$. Thus, the radius $r$ is $7$.

### Phase 3: Constructing the Equation

We have all the ingredients: the center $(1, -1)$ and the radius squared $r^2 = 49$. We use the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$.

Substituting our values, we get $(x - 1)^2 + (y - (-1))^2 = 49$, which simplifies to $(x - 1)^2 + (y + 1)^2 = 49$. Now, let's expand these squares to match the general form:

$$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 49$$

$$x^2 + y^2 - 2x + 2y + 2 = 49$$

$$x^2 + y^2 - 2x + 2y = 47$$

And there it is! The equation of our circle, matching option (3). You have successfully navigated the intersection of lines, the area of a circle, and the expansion of algebraic identities. Keep this momentum going; every problem you solve is a step closer to your goal!

Animated Solution for Mathematics - Circles: The lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle of area 154 sq. units. Then the equation of this circle is

Select Answer:

Visualized Solution

Visualizing the Diameters

Intersection of Diameters

Setting up the Equations

Eliminating $x$

Solving for $y$

Solving for $x$

The Center of the Circle

Using the Area to find Radius

Substituting Area Values

Calculating the Radius

Standard Equation of a Circle

Substituting Center and Radius

Expanding the Equation

Final General Equation

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Circles

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The lines 2x−3y=5 and 3x−4y=7 are diameters of a circle of area 154 sq. units. Then the equation of this circle is