Problem 70 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The equation of the circle passing through the point \((1,0)\) and \((0,1)\) and having the smallest radius is - (a) \(x^{2}+y^{2}-2 x-2 y+1=0\) (b) \(x^{2}+y^{2}-x-y=0\) (c) \(x^{2}+y^{2}+2 x+2 y-7=0\) (d) \(x^{2}+y^{2}+x+y-2=0\)

Step-by-Step Solution

Answer

The equation is (b) \(x^2 + y^2 - x - y = 0\).

Step 1: Recognize the General Equation of a Circle

The general equation of a circle centered at \((h, k)\) with radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). This expands to \(x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0\).

Step 2: Consider Given Points for Circle Equation

We need the circle to pass through the points \((1,0)\) and \((0,1)\). Substitute these points into the general form to create equations.For \((1,0)\):\[1 - 2h + h^2 - 2k + k^2 = r^2\]For \((0,1)\):\[1 - 2k + k^2 - 2h + h^2 = r^2\]These simplify to: 1. \(h^2 + k^2 - 2h = r^2\) 2. \(h^2 + k^2 - 2k = r^2\)

Step 3: Find Simplified Circle Equation

Set the equations equal to each other since they both equal \(r^2\):\[h^2 + k^2 - 2h = h^2 + k^2 - 2k\]Thus, \(-2h = -2k\) implies \(h = k\).

Step 4: Locate the Center on Line y=x

The smallest radius occurs when the center falls exactly in the middle of \((1,0)\) and \((0,1)\). Calculate the midpoint:\[h = k = \frac{1 + 0}{2} = \frac{1}{2}\]So the center is \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Step 5: Calculate Radius of the Circle

Use distance formula to calculate the radius:\[r = \sqrt{\left(\frac{1}{2} - 1\right)^2 + \left(\frac{1}{2} - 0\right)^2} = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2}\]

Step 6: Formulate the Circle Equation

Substitute \(h = \frac{1}{2}, k = \frac{1}{2}, r = \frac{\sqrt{2}}{2}\) back into the equation:\[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \left(\frac{\sqrt{2}}{2}\right)^2\]Expand and simplify:\[x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} = \frac{1}{2}\]\[x^2 + y^2 - x - y = 0\]This matches the circle equation from option (b).

Step 7: Validate and Select Final Answer

Verify that this equation is correct and matches one of the given options. Among the given options, it matches option (b):\(x^2 + y^2 - x - y = 0\).

Key Concepts

Circle Radius

The radius of a circle is a crucial measurement that defines the size of the circle. It is the distance from the center of the circle to any point on its circumference. Understanding how to calculate the circle's radius is essential when you are working with a circle equation.In the context of our problem, the circle is defined such that it passes through the points

• (1, 0)
• (0, 1)

To find the circle with the smallest radius that encompasses these points, we first need to identify the center of this circle precisely on the line \(y = x\). Once we determine this center at \((\frac{1}{2}, \frac{1}{2})\), we use the distance formula to find the radius. The shortest distance from the center to either of the given points will define the smallest circle that can encompass both points. In this exercise, the radius turned out to be \(\frac{\sqrt{2}}{2}\). This precise calculation allows us to formulate the circle equation correctly.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to use algebra and geometry together. In problems where shapes like circles are involved, coordinate geometry provides a powerful way to describe the relationships between points and the shape itself.The equation of a circle is often expressed in the form \((x - h)^2 + (y - k)^2 = r^2\) where

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

In our exercise, we took advantage of coordinate geometry to identify the center point \((h, k)\) that ensures the circle passes through the specific points given: \((1,0)\) and \((0,1)\). By substituting these into our general circle equation and solving, we located the midpoint on line \(y = x\) to achieve the smallest possible circle.This use of coordinate geometry frameworks allows us to tackle complex geometry problems that might otherwise be difficult to visualize or solve using purely geometric methods.

Distance Formula

The distance formula is a mathematical tool used to calculate the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and is essential when working with geometric problems that involve the locations of points.To use the distance formula, you require two points in the form \((x_1, y_1)\) and \((x_2, y_2)\). The formula is expressed as:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In the given exercise, the distance formula is employed to determine the precise radius of the circle. After identifying the center at \((\frac{1}{2}, \frac{1}{2})\), we calculated the distance to point \((1, 0)\), which determined the radius to be \(\frac{\sqrt{2}}{2}\).Using the distance formula here allows for precise calculations that ensure we find the exact size of the circle needed to perfectly intersect the given points. Understanding how to apply the distance formula is a fundamental skill in coordinate geometry and any problems involving distances in the plane.