Problem 58 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The circle passing through \((1,-2)\) and touching the axis of \(x\) at \((3,0)\) also passes through the point (a) \((-5,2)\) (b) \((2,-5)\) (c) \((5,-2)\) (d) \((-2,5)\)

Step-by-Step Solution

Answer

The circle passes through the point (5, -2).

Step 1: Understand the Problem

We are given a problem where a circle passes through the point \((1, -2)\) and touches the x-axis at \((3, 0)\). We need to determine which of the given points the circle also passes through.

Step 2: Establish Known Conditions

Since the circle touches the x-axis at \((3, 0)\), the center of the circle must be \((3, r)\) where \(r\) is the radius. This is because a circle that touches the x-axis has its center vertically at a distance equal to the radius from the point of tangency.

Step 3: Equation of the Circle

Using the center \((3, r)\) and radius \(r\), the equation of the circle is \((x - 3)^2 + (y - r)^2 = r^2\).

Step 4: Substitute Known Point

Substitute the known point \((1, -2)\) into the circle's equation:\((1 - 3)^2 + (-2 - r)^2 = r^2\)which simplifies to \(4 + (r + 2)^2 = r^2\).

Step 5: Simplify and Solve

Expand the equation:\((r + 2)^2 = r^2 - 4\)\(r^2 + 4r + 4 = r^2 - 4\). Cancel \(r^2\) on both sides, and solve:\(4r + 4 = -4\), so \(4r = -8\), then \(r = -2\).

Step 6: Final Equation of Circle

With \(r = -2\), the circle's center is \((3, -2)\) and the equation becomes:\((x - 3)^2 + (y + 2)^2 = 4\).

Step 7: Test Given Points

We test which given point satisfies:\((x - 3)^2 + (y + 2)^2 = 4\).Only the point \((5, -2)\) satisfies:\((5 - 3)^2 + (-2 + 2)^2 = 4\).This simplifies to \(4 + 0 = 4\).

Step 8: Conclusion

Therefore, the correct point is \((5, -2)\).

Key Concepts

Equation of Circle

The equation of a circle in its standard form is an essential element in Circle Geometry. This equation is derived from the general properties of a circle, which is a set of all points in a plane that are equidistant from a given point called the center. The formula for a circle's equation is: \[ (x-h)^2 + (y-k)^2 = r^2 \] where

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

The equation can describe various circles by altering the parameters \( h \), \( k \), and \( r \). In the given problem, we used the conditions that the circle touches the x-axis, which crucially impacted how we determined the circle's center and employed these parameters in our equation.
The circle's equation, \[ (x - 3)^2 + (y - r)^2 = r^2 \], emphasizes how knowing a point and a tangency condition helped derive the equation uniquely suited for that particular scenario.

Radius of Circle

The radius is a pivotal concept in understanding how a circle is situated in a plane. It is the constant distance from the center of the circle to any point it passes, signifying its uniformity and symmetrical structure. When a circle "touches" a line, such as the x-axis given in this problem, the radius provides crucial insights into its geometry.
In the task, the circle is determined to touch the x-axis at \((3, 0)\). This contact point indicates that the center is directly above the line, serving as the highest or lowest vertical point, which equates to the radius. Therefore, knowing the center \( (3, r) \), the radius was initially unknown, but through substitution from the given point \((1, -2)\), we found the radius, \( r = -2 \).
This negative value might appear counter-intuitive, as distance can hardly ever be negative; however, it provides a valid solution in the context of algebraic manipulations in our given scenario, shaping our understanding of the circle's placement and exact configuration.

Circle Properties

Circles hold fascinating properties that can seem almost magical due to their simple yet profound geometric shape. Whether it's the symmetry through their center or the constant distance of every point on the circle from its center, these properties lay the foundation for solving intricate geometric problems.

• A circle always remains in one plane and can be represented by its equation, stressing the mathematical regularity and precision involved.
• The diameter, twice the length of the radius, crosses the center, dividing the circle into two equal parts, establishing another symmetry aspect.
• Tangents form 90-degree angles with the radius at the point of contact with the circle, highlighting a property pivotal in many geometric theorems and real-world applications.

In our problem, the codependency between the circle's center, radius, and equation allowed us to accurately deduce other points the circle passed through, integrating these fundamental circle properties intricately.