Let a circle passing through (2, 0) have its centre at the point (h, k). Let (xc, yc) be the point o

Q: Wait, why can't we just plug $c = 1$ into the equations of the lines immediately?

That is a brilliant question! If you plug $c = 1$ into the second equation, you get $3x + 5y = 1$, which is identical to the first equation. This means the lines become coincident, and they intersect at infinitely many points. We cannot find a unique center that way. By keeping $c$ as a variable and taking the limit *after* solving for the intersection, we are essentially finding the 'limiting position' of the intersection point, which is a standard technique in calculus-based geometry.

Q: Why did we factorize the expression for $x_c$ instead of using L'Hôpital's Rule?

L'Hôpital's Rule is a powerful tool, but in coordinate geometry, algebraic factorization is often cleaner and less prone to calculation errors. When you see a polynomial ratio like $\frac{1 - c^2}{2 + c - 3c^2}$, factorizing reveals the common term $(1 - c)$ that is causing the $0/0$ issue. It is elegant, direct, and keeps your work organized.

Q: Is there a faster way to find the center without doing all this algebra?

You could use Cramer's Rule or determinants to solve for $x_c$ and $y_c$, but the substitution method we used is actually quite efficient here because the equations are linear. The key is to stay calm during the expansion. The algebra might look messy, but notice how the terms cancel out beautifully? That is the hallmark of a well-designed JEE problem.

Question

Let a circle passing through $(2, 0)$ have its centre at the point $(h, k)$. Let $(x_c, y_c)$ be the point of intersection of the lines $3x + 5y = 1$ and $(2 + c)x + 5c^2y = 1$. If $h = \lim_{c 	o 1} x_c$ and $k = \lim_{c 	o 1} y_c$, then the equation of the circle is :

Options:

(A) $25x^2 + 25y^2 - 2x + 2y - 60 = 0$

(B) $5x^2 + 5y^2 - 4x + 2y - 12 = 0$

(C) $5x^2 + 5y^2 - 4x - 2y - 12 = 0$

(D) $25x^2 + 25y^2 - 20x + 2y - 60 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE Advanced. Today, we are not just solving a problem; we are witnessing a beautiful convergence of two distinct worlds: the rigid, structured world of Coordinate Geometry and the fluid, dynamic world of Calculus.

We are tasked with finding the equation of a circle, but the center of this circle is not given to us directly. It is hidden behind a limit, a moving target that we must pin down.

### The Dynamic Intersection

Imagine you are standing on a coordinate plane. You have two lines, $3x + 5y = 1$ and $(2 + c)x + 5c^2y = 1$. As the parameter $c$ changes, the second line shifts and rotates, causing the intersection point $(x_c, y_c)$ to dance across the plane.

To find this intersection, we treat $c$ as a constant for a moment. We have a system of two linear equations. From the first equation, we isolate $5y$:

$$5y = 1 - 3x$$

Now, look at the second equation: $(2 + c)x + c^2(5y) = 1$. By substituting our expression for $5y$, we eliminate $y$ entirely:

$$(2 + c)x + c^2(1 - 3x) = 1$$

Expand this carefully to get $(2 + c)x + c^2 - 3c^2x = 1$. Grouping the $x$ terms gives us $x(2 + c - 3c^2) = 1 - c^2$. Solving for $x_c$, we arrive at:

$$x_c = \frac{1 - c^2}{2 + c - 3c^2}$$

### The Limit of Precision

Now, we enter the realm of Calculus. We need the limit as $c 	o 1$. If you plug $c = 1$ directly into the expression above, you get $0/0$, which indicates a hidden factor of $(1 - c)$ in both the numerator and the denominator.

Let us factorize:

$$1 - c^2 = (1 - c)(1 + c)$$
$$2 + c - 3c^2 = (1 - c)(2 + 3c)$$

When we divide these, the $(1 - c)$ terms cancel out, leaving us with a clean, well-behaved function:

$$x_c = \frac{1 + c}{2 + 3c}$$

Applying the limit $c 	o 1$ is now trivial. We get $h = \frac{1 + 1}{2 + 3(1)} = \frac{2}{5}$.

By substituting this back into our relation $5y_c = 1 - 3x_c$, we find $y_c = \frac{1 - 3(\frac{2}{5})}{5} = \frac{-1/5}{5} = -\frac{1}{25}$. Our center $C$ is at $(\frac{2}{5}, -\frac{1}{25})$.

### The Final Assembly

We have the center. We know the circle passes through $P(2, 0)$. The radius squared, $r^2$, is the squared distance between $C$ and $P$:

$$r^2 = \left(2 - \frac{2}{5}ight)^2 + \left(0 - \left(-\frac{1}{25}ight)ight)^2$$

$$r^2 = \left(\frac{8}{5}ight)^2 + \left(\frac{1}{25}ight)^2 = \frac{64}{25} + \frac{1}{625} = \frac{1600 + 1}{625} = \frac{1601}{625}$$

Finally, we write the equation of the circle in standard form: $(x - h)^2 + (y - k)^2 = r^2$. Substituting our values:

$$\left(x - \frac{2}{5}ight)^2 + \left(y + \frac{1}{25}ight)^2 = \frac{1601}{625}$$

Expanding this and multiplying by $625$ to clear the denominators, we arrive at the final result:

$$25x^2 + 25y^2 - 20x + 2y - 60 = 0$$

This is the equation of our circle. It matches Option 4 perfectly. This is the essence of JEE Advanced mathematics—connecting disparate concepts to reveal a single, **elegant truth**.

Select Answer:

Visualized Solution

Visualizing the Problem

Expressing $5 y$ from Equation 1

Substitution into Equation 2

Isolating $x_{c}$

Factorizing the Expression

Finding the $x$ -coordinate of Center ( $h$ )

Finding $y_{c}$ in terms of $c$

Finding the $y$ -coordinate of Center ( $k$ )

Calculating the Radius Squared ( $r^{2}$ )

Simplifying $r^{2}$

Writing the Circle Equation

Expanding the Terms

Final Simplification

The Final Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Dynamic Intersection

The Limit of Precision

The Final Assembly

Similar Questions

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 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

Intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . Equation of the circle on $A B$ as a diameter is

The points of intersection of the line $a x + b y = 0, (a \neq = b)$ and the circle $x^{2} + y^{2} - 2 x = 0$ are $A (α, 0)$ and $B (1, β)$ . The image of the circle with $A B$ as a diameter in the line $x + y + 2 = 0$ is :

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

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

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

The centre of a circle passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

Lines $5 x + 12 y - 10 = 0$ and $5 x - 12 y - 40 = 0$ touch a circle $C_{1}$ of diameter 6. If the centre of $C_{1}$ lies in the first quadrant, find the equation of the circle $C_{2}$ which is concentric with $C_{1}$ and cuts intercepts of length 8 on these lines.

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

Select Answer:

Visualized Solution

Visualizing the Problem

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5y from Equation 1

Substitution into Equation 2

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

Factorizing the Expression

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate of Center ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yc​ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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

Calculating the Radius Squared ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2