Problem 45 - A Complete Resource Book in Mathematics for JEE Main

Question

The locus of the centre of a circle which passes through the point \((0,0)\) and cuts off a length \(2 b\) from the line \(x\) \(=c\), is (A) \(y^{2}+2 c x=b^{2}+c^{2}\) (B) \(x^{2}+c x=b^{2}+c^{2}\) (C) \(y^{2}+2 c y=b^{2}+c^{2}\) (D) none of these

Step-by-Step Solution

Answer

The correct option is (A): \(y^2 + 2cx = b^2 + c^2\).

Step 1: Understand the Problem

We need to determine the equation of the locus of the center of a circle. The circle passes through the origin (0,0) and cuts a chord of length \(2b\) from the line \(x = c\). The possible answers are: \(y^2 + 2cx = b^2 + c^2\), \(x^2 + cx = b^2 + c^2\), \(y^2 + 2cy = b^2 + c^2\), or none of these.

Step 2: Consider the Circle Equation

Consider a circle with center \((h, k)\) and radius \(r\). The equation for the circle is \((x - h)^2 + (y - k)^2 = r^2\). The circle passes through the origin, so substituting \((0, 0)\) gives \(h^2 + k^2 = r^2\).

Step 3: Determine Chord Condition

The given line, \(x = c\), cuts off a chord of length \(2b\). Using the geometry of chords and circles, the distance \(d\) of the center of the circle from the line is \(\sqrt{r^2 - b^2}\). Here, the line \(x = c\) translates to \(\|h - c\| = \sqrt{r^2 - b^2}\).

Step 4: Solve for Locus Equation

We have two equations: \(h^2 + k^2 = r^2\) and \((h-c)^2 = r^2 - b^2\). Expanding and simplifying the second equation gives: \(h^2 - 2hc + c^2 = r^2 - b^2\). Subtract this from the first equation, we have: \(k^2 = b^2 + c^2 - 2hc\). So, \(k^2 + 2hc = b^2 + c^2\). Assuming \(x = h\) and \(y = k\), we get \(y^2 + 2cx = b^2 + c^2\).

Step 5: Match with Given Options

Comparing with the provided options, the derived equation for the locus \(y^2 + 2cx = b^2 + c^2\) matches with option (A).

Key Concepts

Circle Equation

The equation of a circle is a foundational concept in geometry. It describes all the points that are equidistant from a fixed point called the center. If the center of the circle is \( (h, k) \) and the radius is \( r \), the circle's equation is given by: \((x - h)^2 + (y - k)^2 = r^2\).
This equation essentially tells you that for any point \( (x, y) \) on the circle, the distance from \( (h, k) \) is always \( r \). When a specific point, such as the origin \( (0,0) \), is on the circle, you can substitute these coordinates into the circle equation. This process reveals the relationship between the circle's center and its radius.
- **Example:** Substituting \( (0,0) \) into the circle's equation yields \( h^2 + k^2 = r^2 \). This tells us that even though we don’t know the exact radius, we know it depends on \( h \) and \( k \).

Chord Conditions

When we talk about a chord in a circle, we refer to a straight line connecting two points on the circle. The length of a chord is determined by its distance from the center and the angle it subtends. In our problem, the length of the chord cut from the line \( x = c \) is given as \( 2b \).
To find out how chords relate to the circle's properties, we use the perpendicular distance from the circle’s center to the chord. The formula for this distance \( d \) when the chord’s length is given, is: \(\sqrt{r^2 - b^2}\). This essentially arises from geometry, where the Pythagorean theorem helps express relations in right triangles formed by the radius, the distance \( d \), and half the chord.
- **Application:** In the context of our problem, if the chord lies along the line \( x = c \), then the perpendicular distance \( |h - c| = \sqrt{r^2 - b^2} \), expressing how far the circle's center is from this line.

Geometry of Circles

The geometry of circles explores the characteristics and relationships within a circle. It focuses on components like the radius, diameter, and various lines like tangents and chords. Understanding these relationships is crucial in solving problems involving circles.
The equations derived from circle geometry can provide insights, such as finding the locus of points. The locus is essentially a set of points that satisfy specific conditions. In our exercise, the problem is about the locus of the center of a circle.
- **Locus of Points:** By using the circle's properties and given conditions, like passing through a point or cutting a specific chord, we derived an equation for the locus: \((y^2 + 2cx = b^2 + c^2)\). This represents all possible locations for the circle's center satisfying the conditions of passing through the origin and intersecting a line to from a specific chord length.