Problem 94 - A Complete Resource Book in Mathematics for JEE Main

Question

The pole of the chord of the circle \(x^{2}+y^{2}=16\) which is bisected at the point \((-2,3)\), with respect to the circle is (A) \(\left(\frac{-32}{13}, \frac{48}{13}\right)\) (B) \(\left(\frac{32}{13}, \frac{48}{13}\right)\) (C) \(\left(\frac{-32}{13}, \frac{-48}{13}\right)\) (D) none of these

Step-by-Step Solution

Answer

None of these options are correct.

Step 1: Identify the Circle Equation

The equation of the given circle is \(x^2 + y^2 = 16\). The center of this circle \((h, k)\) is \((0, 0)\) and the radius is 4.

Step 2: Understanding the Concept

The pole of a chord is found using the formula: \(x_1x + y_1y = a^2\) where the chord is bisected at \((x_1, y_1)\) and \(a^2\) is the radius squared of the circle. Here, the chord is bisected at \((-2, 3)\) and \(a^2 = 16\).

Step 3: Write the Equation of the Chord

Using the formula \(x_1x + y_1y = a^2\), substitute \(x_1 = -2\), \(y_1 = 3\), and \(a^2 = 16\). The equation of the chord is:\[-2x + 3y = 16\]

Step 4: Convert to Standard Form

Rewriting the chord equation in the standard form, we isolate terms on one side:\[-2x + 3y - 16 = 0\]

Step 5: Find the Pole Point

The pole of the chord \(Ax + By + C = 0\) with respect to a circle \(x^2 + y^2 = a^2\) is given by:\[(x', y') = \left(\frac{-a^2 A}{C}, \frac{-a^2 B}{C}\right)\]Substitute \(A = -2\), \(B = 3\), \(C = -16\), and \(a^2 = 16\) into the formula.

Step 6: Calculate Coordinates of the Pole

Compute \((x', y')\):\[x' = \frac{-16(-2)}{-16} = \frac{32}{16} = 2\]\[y' = \frac{-16(3)}{-16} = \frac{-48}{16} = -3\]

Key Concepts

Circle Equation

In coordinate geometry, a circle is often described using its equation in the standard form. This makes it easy to identify crucial properties such as the circle's center and radius. The general equation for a circle with a center at

• ewlineThe standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \).
• When we see an equation like \( x^2 + y^2 = 16 \), it tells us a few things:
  • The circle is centered at the origin, \((0, 0)\).
  • The radius is 4, since \( r^2 = 16 \), and taking the square root gives us 4.

These foundational properties enable us to expand into more detailed calculations, such as determining other geometric features like poles and tangents related to the circle.

Pole of a Chord

The concept of the pole of a chord is intriguing in geometric constructions. In the context of a circle, a chord is essentially a line segment whose endpoints lie on the circle. The pole of a chord is a unique point associated with this chord, found using a specific formula:

• The chord is considered to be bisected at a specific point, in this task, it is \((-2, 3)\).
• This reflects how a chord divides into two equal segments.

To find the pole, we use the chord equation given within the circle's context. By leveraging the formula \( x_1 x + y_1 y = a^2 \), where \( (x_1, y_1) \) represents where the chord is bisected, we integrate these values. In our case, substituting

• \( x_1 = -2 \)
• \( y_1 = 3 \)
• \( a^2 = 16 \)

Into the chord equation gives us a clearer foundation to identify the specific line equation, eventually leading us to a better understanding and calculation of the pole.

Coordinate Calculation

Coordinate calculations are crucial when dealing with geometric positions, especially for determining points like the pole of a chord. Once we have the standard form of the chord line, \[ -2x + 3y - 16 = 0 \], identifying the pole involves substituting into a specific formula.

• The formula used here is \((x', y') = \left( \frac{-a^2 A}{C}, \frac{-a^2 B}{C} \right)\), which revolves around the coefficients of the line equation.
• By substituting:
  • \( A = -2 \) (coefficient of \( x \))
  • \( B = 3 \) (coefficient of \( y \))
  • \( C = -16 \) (the constant term)
  • \( a^2 = 16 \)
  We proceed to calculate as follows:
• \( x' = \frac{-16(-2)}{-16} = 2 \)
• \( y' = \frac{-16(3)}{-16} = -3 \)

These calculations provide the precise coordinates \((2, -3)\), marking the pole location relative to the circle. Practicing these steps improves understanding and accuracy in coordinate geometry.