Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Circles are drawn through the point \((-5,0)\) to cut the \(x\)-axis on the positive side and making an intercept of 10 units on the \(x\)-axis. The equation of the locus of the centre of these circles is (A) \(x+y=0\) (B) \(x-\bar{y}=0\) (C) \(x=0\) (D) \(y=0\)

Step-by-Step Solution

Answer

(A) The equation of the locus is \(x + y = 0\).

Step 1: Understanding Circle Properties

A circle's general equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle, and \(r\) is the radius.

Step 2: Intercept Property on X-axis

The circle intersects the x-axis at two points. These intercepts have the form \((a, 0)\) and \((b, 0)\). The length of this intercept, \(\lvert a - b \rvert\), is given as 10 units.

Step 3: Interpreting Point on X-axis and Radius

Given that one point on the x-axis is positive and has a distance of 10 units, we can infer that \(a = b - 10\) on the x-axis.

Step 4: Calculate Center's Coordinates Relationship

Since the point \((-5, 0)\) lies on the circle, any circle must have its center equidistant from this point and the line segment joining \(a\) and \(b\). Therefore, the midpoint of the x-intercepts, \(\left(\frac{a+b}{2}, 0\right)\), leads to the center line equation \(h + k = 0\).

Step 5: Finding Locus Equation

Using the derived relationship \(k = -h\), the equation of the locus of the center is derived: \(x + y = 0\).

Key Concepts

Circle Equation

The circle equation is fundamental in understanding circles in coordinate geometry. The general form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where the pair \((h, k)\) represents the center of the circle while \(r\) denotes the radius. This equation defines all points \((x, y)\) that are equidistant \(r\) from the center \((h, k)\). It essentially captures every point lying on the edge of the circle. Understanding this basic equation allows us to explore further properties like intercepts and other geometrical relationships associated with circles.

Intercept on x-axis

Intercepts on the x-axis occur where the circle crosses the x-axis. For a circle with center \((h, k)\), these points are often given as \((a, 0)\) and \((b, 0)\). The distance between these two intercepts is the x-axis intercept, which provides valuable information about the circle's relation to the x-axis.
In the problem, it's given that the total intercept is 10 units long. This simply means that the distance \(|a-b|\) equals 10. Such intercepts are significant as they help us define measurements like the circle's diameter or radius, depending on additional given conditions.

Center of Circle

Determining the center of a circle is crucial for identifying a circle's position in the plane. The center, denoted by \((h, k)\), is determined using known points and conditions. Given a point through which all circles pass, \((-5, 0)\) in this case, the center must be equally distanced from this point as from any x-axis intercepts.
Usually, the center lies on the axis of symmetry for the x-intercepts, which can be calculated as the midpoint \(\left(\frac{a+b}{2}, 0\right)\). In the example provided, the derived relation \(h + k = 0\) gives us the positions where centers of all satisfying circles could lie, by linking it through a geometric property.

Locus Equation

The concept of a locus involves determining a set of points that share a common property or satisfy certain conditions. In this context, the locus equation refers to the path traced by the centers of circles fitting the problem's criteria.
Through geometrical reasoning, considering circles that pass through \((-5, 0)\) and exhibit 10-unit x-intercepts, we deduce that the center of each such circle satisfies \(k = -h\). This means each center lies on the line characterized by the equation \(x + y = 0\). This line serves as the locus of the center of all these circles, effectively mapping out their possible locations in the coordinate plane.