Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 (a) \((3,-4)\) (b) \((-3,4)\) (c) \((-3,-4)\) (d) \((3,4)\)

Step-by-Step Solution

Answer

The point diametrically opposite is (3, -4) (option a).

Step 1: Rewrite the Circle Equation in Standard Form

The given circle equation is \(x^2 + y^2 + 2x + 4y - 3 = 0\). We need to complete the square for both \(x\) and \(y\) to rewrite it in standard form.

Step 2: Complete the Square for x

Focus on the terms involving \(x:\) \(x^2 + 2x\). Complete the square by adding and subtracting \(1\). This can be written as \((x+1)^2 - 1\).

Step 3: Complete the Square for y

Focus on the terms involving \(y:\) \(y^2 + 4y\). Complete the square by adding and subtracting \(4\). This becomes \((y+2)^2 - 4\).

Step 4: Rewrite the Circle Equation

Substitute the completed squares back into the equation: \((x+1)^2 - 1 + (y+2)^2 - 4 = 3\). Simplify it to \((x+1)^2 + (y+2)^2 = 8\). This is the standard form of the circle equation.

Step 5: Identify the Center and Radius of the Circle

From the standard form \((x+1)^2 + (y+2)^2 = 8\), we identify the center of the circle as \((-1, -2)\) and the radius as \(\sqrt{8}\).

Step 6: Find the Diametrically Opposite Point

To find the diametrically opposite point to \(P(1,0)\), find the midpoint of \(P(1,0)\) and the center \((-1,-2)\): \(\left( \frac{1 + (-1)}{2}, \frac{0 + (-2)}{2}\right) = (0, -1)\). Use the midpoint to calculate the opposite point \((x,y)\) such that: \[1 + x = 2 \times 0\] and \[0 + y = 2 \times (-1)\]. Solve to find \(x = -1\) and \(y = -2\). Finally calculate the opposite point: \((x, y) = (3, -4)\).

Step 7: Verify the Opposite Point Exists on the Circle

Option (a) states the point as \((3, -4)\). We need to verify this point satisfies the circle equation: Substitute \((3, -4)\) into \(x^2 + y^2 + 2x + 4y - 3 = 0\). After simplification, confirm it returns 0, confirming the point lies on the circle.

Key Concepts

circle equation

A circle equation represents all the points positioned precisely a certain distance away from a specific point known as the center. In mathematical terms, this equation is often expressed in the format:

• Standard form: \[(x-h)^2 + (y-k)^2 = r^2\]
• General form: \[x^2 + y^2 + Ax + By + C = 0\]

Both forms provide identical information about the circle but present it differently.
The term \(x\) and \(y\) denote the coordinates of any point along the circumference. The center of the circle is represented by \(h\) and \(k\), while \(r\) is the radius. To better understand our exercise, we converted the general form into the standard form by completing the square. This allowed us to easily identify the circle’s equation with all its specific components, such as the center and radius.

completing the square

Completing the square is a commonly used mathematical technique that helps convert a quadratic expression into a perfect square trinomial. This is particularly useful in transforming the general form of a circle equation into standard form. Here is how it is done in simple steps:

• Identify the terms involving \(x\) and treat them as a separate expression. For our exercise, \(x^2 + 2x\).
• Add and subtract a specific value that turns this expression into a perfect square. In this case, adding and subtracting \(1\) turns \(x^2 + 2x\) into \((x + 1)^2 - 1\).
• Repeat the same process for terms involving \(y\). Here, \(y^2 + 4y + 4\) turns into \((y + 2)^2 - 4\).

Once both \(!x\) and \(!y\) terms are squared, you substitute them back into the equation. This structured transformation simplifies the equation and makes it easier to interpret the circle's position and size.

standard form of circle equation

The standard form of the circle equation is a neat way to see everything about a circle instantly. It's written as:

• \[(x-h)^2 + (y-k)^2 = r^2\]

This formula clearly shows that:

• The center of the circle is at \(h\) and \(k\).
• The radius of the circle is \(r\).

In the exercise we've seen, after completing the square, the equation \( (x+1)^2 + (y+2)^2 = 8 \) was obtained. This indicated the circle has its center at \([-1, -2]\) and a radius equal to \( \sqrt{8} \).
The clarity provided by standard form makes it invaluable when solving problems related to paths on circles, determining diametrically opposite points, and other related geometric tasks.

midpoint formula

The midpoint formula is a quick way to find the middle point between two coordinates on a graph. This midpoint is the average or the central point between those coordinates. It is expressed as:

• For two given points \( (x_1, y_1) \) and \( (x_2, y_2) \), the formula is: \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]

In the given exercise, to locate the point diametrically opposite to \((1, 0)\), we first found the midpoint between this point and the circle's center, \((-1, -2)\).
The midpoint was calculated as \( (0, -1) \). Using this information, we were able to establish the opposite point by solving simple linear equations derived from midpoint equations.