Problem 42 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If one of the diameters of the circle, given by the equation, \(x^{2}+y^{2}-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\), then the radius of \(S\) is: (a) 5 (b) 10 (c) \(5 \sqrt{2}\) (d) \(5 \sqrt{3}\)

Step-by-Step Solution

Answer

The radius of circle S is \(5 \sqrt{2}\).

Step 1: Identify the Equation of the Circle

The given circle's equation is \(x^{2}+y^{2}-4x+6y-12=0\). To find the center and radius, rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\).

Step 2: Complete the Square

Rewrite the equation by completing the square for both \(x\) and \(y\).\[\begin{align*}x^2 - 4x & : \quad (x-2)^2 = x^2 - 4x + 4,\y^2 + 6y & : \quad (y+3)^2 = y^2 + 6y + 9.\end{align*}\]The equation becomes:\((x-2)^2 + (y+3)^2 = 12+4+9 = 25\).

Step 3: Identify Circle Properties

From the equation \((x-2)^2 + (y+3)^2 = 25\), the center \((h, k)\) is at \((2, -3)\) and the radius \(r\) is \(5\, (\sqrt{25})\).

Step 4: Use Diameter as Chord in Circle S

The circle S has center \((-3, 2)\). Since the diameter equals chord, it passes through \((2, -3)\) and another point symmetrical across the center, say \(P\), making \(d\) symmetrical.

Step 5: Calculate Distance for Symmetrical Point

Find the distance between \((2, -3)\) and the center of \(S\), then mirror to find \(P\). Using distance formula: \[d = \sqrt{(2-(-3))^2 + (-3-2)^2} = \sqrt{25 + 25} = 5\sqrt{2}.\] This is half the chord length, so the diameter is \(2 \times 5\sqrt{2} = 10\sqrt{2}\).

Step 6: Determine the Radius of Circle S

Since the calculated diameter of circle \(S\) is \(10\sqrt{2}\), the radius is half of the diameter, thus \(5\sqrt{2}\).

Key Concepts

Equation of a Circle

In geometry, the equation of a circle is a crucial concept. It defines all the points that make up the shape of the circle on a coordinate plane. A circle is defined by its center \( (h, k) \) and radius \( r \). The general standard form of a circle's equation is given as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
This equation tells us that any point \( (x, y) \) on the circle is exactly \( r \) units away from the center \( (h, k) \).

• \( (x - h)^2 \) and \( (y - k)^2 \) represent the squared distances from any point on the circle to the center, along the x and y axes respectively.
• \( r^2 \) is the square of the radius of the circle.

Understanding this equation allows us to graph the circle and to discern its geometric properties from an algebraic expression.

Completing the Square

Completing the square is a technique used to rewrite a quadratic expression in a form that reveals important geometric properties. It is particularly useful for circles because it helps transform the general form of a circle's equation into the standard form.
Consider a quadratic expression such as \( x^2 - 4x \). The goal is to turn this into a perfect square trinomial. To do this:

• Take half of the coefficient of \( x \), square it, and add it to \( x^2 - 4x \).
• For \( x^2 - 4x \), half of \(-4\) is \(-2\), and \( (-2)^2 = 4 \).
• The expression becomes \( (x - 2)^2 - 4 \).

Repeat this process for \( y \) terms to rewrite each as a square.
This procedure helps simplify and rearrange the equation to readily identify the center and radius of the circle.

Radius Calculation

Calculating the radius of a circle from its equation is straightforward once it's in standard form. Once you have:
\[ (x - 2)^2 + (y + 3)^2 = 25 \]
In this form, \( r^2 = 25 \). Therefore, the radius \( r \) is the square root of \( 25 \), which is \( 5 \).

• Identify the term on the right side of the equation, which represents the square of the radius.
• Take the square root of this value to find the radius.

The radius determines the size of the circle and is an essential measurement for understanding the circle's properties.

Diameter as a Chord

In circle geometry, any diameter of a circle can be considered a special type of chord. This becomes significant when a diameter of one circle becomes a chord of another circle. In this scenario, the original circle's diameter runs straight through its center, connecting two points on its edge.
For a circle \( S \) where this diameter acts as a chord, considerations include:

• Identifying the endpoints of the diameter, which are points on both circles.
• Understanding that the point of symmetry relative to the center of circle \( S \) means the diameter (or chord) divides the circle into two equal halves.
• The length of this chord can be calculated using the midpoint formula and the distance formula to confirm its position in space.

Knowing this helps in assessing the length and relevance of the diameters when they function as chords in larger circles.