Problem 124 - A Complete Resource Book in Mathematics for JEE Main

Question

The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^{2}+y^{2}+2 x+4 y-3=0\) is |2008| (A) \((3,-4)\) (B) \((-3,4)\) (C) \((-3,-4)\) (D) \((3,4)\)

Step-by-Step Solution

Answer

The point diametrically opposite to \(P(1,0)\) is \((-3,-4)\) (C).

Step 1: Identify the given equation

The circle's equation is given as \(x^{2} + y^{2} + 2x + 4y - 3 = 0\). This is in the general form \(x^2 + y^2 + 2gx + 2fy + c = 0\), where \(g = 1\), \(f = 2\), and \(c = -3\).

Step 2: Rewrite the equation in the standard form

To find the center of the circle, rewrite the equation in standard form by completing the square. The equation becomes: \((x + 1)^2 + (y + 2)^2 = 8\), indicating the center is \((-1, -2)\) and the radius is \(\sqrt{8}\).

Step 3: Determine the diametrically opposite point

The point \(P(1,0)\) and its diametrically opposite point \(Q(x,y)\) must lie on a diameter, meaning the midpoint between \(P\) and \(Q\) is the center of the circle. The midpoint \(M((1+x)/2, (0+y)/2)\) should equal \((-1, -2)\).

Step 4: Set midpoint coordinates equal to the circle's center

Set the coordinates of midpoint \(M\) equal to the coordinates of the circle's center:1. \((1+x)/2 = -1\)2. \((0+y)/2 = -2\)

Step 5: Solve for the coordinates of point Q

Solve the equations from Step 4:1. \((1 + x)/2 = -1\) implies \(1 + x = -2\). Solving gives \(x = -3\).2. \((0 + y)/2 = -2\) implies \(y = -4\).

Step 6: Verify the solution

The solution \((-3, -4)\) satisfies the conditions for being diametrically opposite the point \((1, 0)\). The solution must lie on the circle and be symmetrical with respect to the center.

Key Concepts

Equation of a Circle

A circle's equation beautifully captures its geometry in an algebraic form. The standard form of a circle's equation is given by \[(x - h)^2 + (y - k)^2 = r^2,\]where

• \((h, k)\) is the center of the circle,
• \(r\) is the radius.

In many problems, the equation might initially appear in the general form:\[x^2 + y^2 + 2gx + 2fy + c = 0.\]To identify the center and radius, you must transform this into the standard form. This is done by completing the square for both \(x\) and \(y\). Only then can you see the clear picture of the circle’s dimensions and position within the coordinate plane. This allows you to solve problems involving circles accurately.One practical application of knowing the equation of a circle is to find specific points related to it, such as diametrically opposite points.

Diametrically Opposite Points

When two points lie diametrically opposite to each other on a circle, they are at opposite ends of a diameter. This means that together, they pass through the center of the circle. If you know one point and need to find the one diametrically opposite to it:

• First, determine the center of the circle using the equation.
• Then, use the midpoint formula, which states that the midpoint of the diameter (or line segment between the two points) must be the center of the circle.

Given a point \(P(x_1, y_1)\), to find the diametrically opposite point \(Q(x_2, y_2)\), utilize the midpoint formula:

• \((x_1 + x_2) / 2 = h\)
• \((y_1 + y_2) / 2 = k\)

Where \((h, k)\) is the center of the circle. Solving these equations will give you the coordinates of the point \(Q\), ensuring it is precisely at the opposite end of the circle.

Completing the Square

The technique of completing the square is a powerful tool in algebra, essential for converting a quadratic polynomial into a more manageable form. It is especially useful when dealing with circle equations in general form. The process involves:

• Identifying the quadratic terms associated with both \(x\) and \(y\).
• Rewriting these in a form where they become easy to square, usually taking half of the linear coefficient, squaring it, and then adding and subtracting this square within the equation.

Take for example the equation \(x^2 + 2ax\) which can be rewritten by completing the square as:\[(x + a)^2 - a^2.\]For a circle’s equation like \(x^2 + y^2 + 2gx + 2fy + c = 0\), completing the square allows us to separate the \(x\) and \(y\) groups, revealing both the circle's center and its radius. Through this method, the algebra simplifies, revealing the geometry lying underneath. Understanding this process not only aids in geometry problems but also provides a deeper grasp of algebraic transformation techniques.