Question

A circle \( S = x^2 + y^2 - 16 = 0 \) intersects another circle \( S' = 0 \) of radius 5 units such that their common chord is of maximum length. If the slope of that chord is \( \dfrac{3}{4} \), then the centre of such a circle \( S' = 0 \) is:

• A. \( \left( \dfrac{9}{5}, \dfrac{12}{5} \right) \)
• B. \( \left( \dfrac{9}{5}, -\dfrac{12}{5} \right) \)
• C. \( \left( -\dfrac{9}{5}, \dfrac{12}{5} \right) \)
• D. \( \left( \dfrac{3}{5}, \dfrac{4}{5} \right) \)

Hint:

When the common chord of two intersecting circles is of maximum length, the line joining the centers is perpendicular to the chord. Use this perpendicularity to find the relationship between the centers.

Answer 1

The Correct Option is C

Step 1: Given Circle \( S \)
Equation: \( x^2 + y^2 - 16 = 0 \Rightarrow \) center = \( (0, 0) \), radius = \( 4 \)
Step 2: Common chord of maximum length
The maximum length of common chord between two circles occurs when the line joining their centers is perpendicular to the chord. So, the direction of the line joining centers is perpendicular to the chord.
Given slope of chord = \( \dfrac{3}{4} \Rightarrow \) slope of line joining centers = \( -\dfrac{4}{3} \)
Step 3: Geometry of centers
Let center of \( S' \) be \( (x, y) \). The center of \( S \) is at origin \( (0,0) \). So,
\[ \text{slope of line joining centers} = \dfrac{y - 0}{x - 0} = \dfrac{y}{x} = -\dfrac{4}{3} \Rightarrow y = -\dfrac{4}{3}x \tag{1} \] Step 4: Distance between centers
Let the distance between the centers be \( d \). Since the chord is of maximum length, the distance between centers is equal to the difference of radii (as one circle lies outside the other, touching through chord).
Radius of \( S = 4 \), Radius of \( S' = 5 \)
\[ \text{Distance between centers} = \sqrt{x^2 + y^2} = \sqrt{x^2 + \left( -\dfrac{4}{3}x \right)^2} = \sqrt{x^2 + \dfrac{16}{9}x^2} = \sqrt{ \dfrac{25}{9}x^2 } = \dfrac{5}{3}|x| \] Set this equal to distance between centers = \( 4 + 5 = 9 \Rightarrow \dfrac{5}{3}|x| = 9 \Rightarrow |x| = \dfrac{27}{5} \)
We test the option with \( x = -\dfrac{9}{5} \), so that:
\[ y = -\dfrac{4}{3} \cdot \left( -\dfrac{9}{5} \right) = \dfrac{12}{5} \Rightarrow \text{Center of } S' = \left( -\dfrac{9}{5}, \dfrac{12}{5} \right) \] \[ \boxed{ \left( -\dfrac{9}{5}, \dfrac{12}{5} \right) } \]