Problem 133 - A Complete Resource Book in Mathematics for JEE Main 2017

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) 10 (B) \(5 \sqrt{2}\) (C) \(5 \sqrt{3}\) (D) 5

Step-by-Step Solution

Answer

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

Step 1: Identify the Center and Radius of the Given Circle

The equation of the circle is given as \(x^{2}+y^{2}-4x+6y-12=0\). We can rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\) by completing the square. This helps us to identify the center \(h, k\) and radius \(r\).

Step 2: Complete the Square

First, complete the square for both the \(x\) and \(y\) terms in the equation. For \(x^2 - 4x\), add and subtract \(4\) (since \((-4/2)^2 = 4\)). For \(y^2 + 6y\), add and subtract \(9\) (since \((6/2)^2 = 9\)). This gives us: \((x-2)^2 - 4 + (y+3)^2 - 9 - 12 = 0\).

Step 3: Simplify the Equation

Rearrange the terms: \((x-2)^2 + (y+3)^2 = 25\). This is the standard form of the circle's equation. Thus, the center is \((2, -3)\) and the radius is \(5\).

Step 4: Equation of the Diameter

The equation \( (x-2)^2 + (y+3)^2 = 25 \) provides a circle with diameter \((1,0)\) by using its coefficient structure. Determine that the midpoint, as part of the larger circle, divides this segment into two equal parts based on its central line which outlines the primary circle \(S\).

Step 5: Distance between Center of S and Given Circle

Calculate the distance between the center of circle \(S\) \((-3,2)\) and the center of the given circle \((2,-3)\) using the distance formula: \( \sqrt{(2 + 3)^2 + (-3 - 2)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}\).

Step 6: Determine the Radius of Circle S

Since the given circle's diameter is the chord of the circle \(S\), and the orientation places it equidistant across the diameter center-to-center, the circle \(S\) balances the mid-segment distance across. It completes the successful match-up, allowing the solution of \(5\sqrt{2}\).

Key Concepts

Equation of a Circle

The equation of a circle in its standard form is typically expressed as \((x-h)^2 + (y-k)^2 = r^2\). In this equation:

• \(h\) represents the x-coordinate of the circle's center.
• \(k\) represents the y-coordinate of the circle's center.
• \(r\) is the radius of the circle.

This form clearly shows both the center and the radius, making it easier to visualize the circle's position and size in a coordinate plane. To transform any general circle equation into this standard form, we often use a technique called "completing the square." This transformation helps identify the circle's key elements by restructuring the equation. Once it's in standard form, it becomes straightforward to read off the center and radius.

Completing the Square

Completing the square is a method used to rewrite a quadratic expression into a perfect square trinomial. This is helpful in converting circle equations into their standard form. To complete the square:

• Take the coefficient of the linear term (the term with x or y), divide it by 2, and square it.
• Add and subtract this new value inside the equation to form a perfect trinomial square.

Consider the term \(x^2 - 4x\) in the given circle equation. Divide \(-4\) by 2, giving \(-2\), and then square it to get \(4\). Thus:\[(x^2 - 4x) \Rightarrow (x-2)^2 - 4\]Similarly for \(y^2 + 6y\), divide \(6\) by 2 to get \(3\), then square it to get \(9\). Thus:\[(y^2 + 6y) \Rightarrow (y+3)^2 - 9\]By completing the square, the equation \(x^2 + y^2 - 4x + 6y - 12 = 0\) becomes \((x-2)^2 + (y+3)^2 = 25\), revealing the circle's center and radius easily.

Distance Formula

In geometry, the distance formula helps calculate the distance between two points on a coordinate plane. It's derived from the Pythagorean Theorem. The formula is:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Where:

• \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of the two points.

Applying the distance formula, you can find the distance between, for example, the centers of two circles. In the exercise, to determine the distance between the center of the given circle \((2, -3)\) and the center of circle \(S\) at \((-3, 2)\), the formula becomes:\[\sqrt{(2 + 3)^2 + (-3 - 2)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}\]Knowing this distance can help solve for other circle properties, like determining if a diameter of one circle is a chord of another.

Circle Center and Radius

The center and radius are crucial to identifying a circle's properties. The center, given by \((h, k)\), dictates the circle's location in the plane, while the radius \(r\) states its size.

• A change in the center affects the circle’s position without altering its size.
• The radius determines the distance from any point on the circle's boundary to its center.

From the exercise's equation \((x-2)^2 + (y+3)^2 = 25\), the center is clearly \((2, -3)\), and the radius is found by taking the square root of 25, resulting in \(r = 5\). For circle \(S\), with the center at \((-3, 2)\), determining its radius involves understanding its relation to the given circle's diameter, which acts as a chord for \(S\). Calculating the distances and orientating them appropriately, it's concluded that the radius of circle \(S\) balances effectively, resulting in \(5\sqrt{2}\). Thus, correctly identifying these components is essential for comprehending the spatial setup of circles.