Problem 121 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{P}\) be the point on the parabola, \(\mathrm{y}^{2}=8 \mathrm{x}\) which is at a minimum distance from the centre \(\mathrm{C}\) of the circle, \(\mathrm{x}^{2}+(\mathrm{y}+6)^{2}=1\). Then the equation of the circle, passing through \(\mathrm{C}\) and having its centre at \(\mathrm{P}\) is: (a) \(x^{2}+y^{2}-\frac{x}{4}+2 y-24=0\) (b) \(x^{2}+y^{2}-4 x+9 y+18=0\) (c) \(x^{2}+y^{2}-4 x+8 y+12=0\) (d) \(x^{2}+y^{2}-x+4 y-12=0\)

Step-by-Step Solution

Answer

The circle's equation passing through C with center at P is (a) \(x^2 + y^2 - \frac{x}{4} + 2y - 24 = 0\).

Step 1: Understand the Parabola Equation

The given parabola is described by the equation \(y^2 = 8x\). This represents a standard parabola that opens to the right.

Step 2: Understand the Circle Equation

The given equation \(x^2 + (y+6)^2 = 1\) describes a circle with center at \(C(0, -6)\) and a radius of 1.

Step 3: Find the Point P on the Parabola

To find point \(P(h, k)\) on the parabola that is at a minimum distance from point \(C(0, -6)\), express \(k^2 = 8h\). The distance from \(P\) to \(C\) is \(\sqrt{(h-0)^2 + (k+6)^2}\). Use this in the step where we minimize this distance.

Step 4: Minimize the Distance

To minimize the distance \(\sqrt{h^2 + (k+6)^2}\), differentiate the distance with respect to \(k\) and set the derivative to zero. Solving gives \(k = -2\), and substituting in the parabola equation gives \(h = \frac{1}{2}\). Thus, \(P\) is \(\left(\frac{1}{2}, -2\right)\).

Step 5: Equation of the Circle through Points C and P

Now we need the equation of the circle with center \(\left(\frac{1}{2}, -2\right)\) and passing through \(C(0, -6)\). Calculate the radius as the distance between \(P\) and \(C\), which is \(\sqrt{\left(\frac{1}{2}-0\right)^2 + (-2 + 6)^2} = \frac{9}{2}\).

Step 6: Derive the Circle Equation

The circle's equation with center \(\left(\frac{1}{2}, -2\right)\) and radius \(\frac{9}{2}\) is \((x - \frac{1}{2})^2 + (y + 2)^2 = \left(\frac{9}{2}\right)^2\). Expand and simplify this to find the equation of the circle, which matches option (a).

Key Concepts

Parabola Equation

The equation of a parabola is fundamental in understanding its geometric properties. A parabola is a symmetric curve and is defined by its equation. In our example, we have the parabola given by the equation \(y^2 = 8x\). Here, the equation is in its standard form, \(y^2 = 4ax\), where \(a\) is a constant that indicates the distance from the vertex to the focus. For this parabola, \(a = 2\), showing that it is positioned with its vertex at the origin \((0,0)\) and opens to the right.
Parabolas have several key components:

• **Vertex**: The point \((0,0)\) where the parabola changes direction.
• **Focus**: Located at \((2,0)\) for this particular parabola, which is \(a\) units along the direction the parabola opens.
• **Directrix**: A line perpendicular to the axis of symmetry, situated at \(x = -2\).

In problems involving optimization or finding distances, pinpointing the vertex, focus, and understanding the opening direction is crucial. In our case, the point \(P\) is on this parabola and we leverage this equation to find the relation between \(h\) and \(k\), where \(P(h, k)\) is such that \(k^2 = 8h\). This relationship simplifies solving the minimizing distance function.

Circle Equation

A circle is defined by all points equidistant from a fixed central point. Its standard equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. For the circle in our exercise, the equation given is \(x^2 + (y+6)^2 = 1\). This tells us:

• **Center**: \((0, -6)\), indicating the circle's midpoint.
• **Radius**: \(1\), the distance from the center to any point on the circle's perimeter.

This circle’s properties play a pivotal role when calculating distances between it and other geometric figures, like the parabola. Additionally, we need these to establish another circle with a specified center and passing through a designated point. Ultimately, the point \(P\) on the parabola that minimizes distance to \(C\) becomes our new circle's center.

Minimizing Distance

The goal of minimizing distance in this context is to identify the point on the parabola closest to the circle’s center. Here, the point \(C\) \((0, -6)\) serves as a reference. To find minimum distance, we calculate the distance formula \(\sqrt{h^2 + (k+6)^2}\) between \(C\) and potential points \(P\) on the parabola. The essential process involves:

• **Defining the function**: Represent the distance in terms of parabola's known parameters.
• **Differentiating**: Find the derivative of the distance with respect to the variable you’re optimizing over, here it’s \(k\).
• **Solving for critical points**: Set the derivative equal to zero to locate potential points of minimum distance.

Upon solving, we ascertain that the minimum distance occurs at \(P\left(\frac{1}{2}, -2\right)\). This point is then used to identify the properties of a new circle, which meets the criteria set in our original problem. Discovering the minimal point makes it possible to draw the circle with center at \(P\) and passing through \(C\).