Question

Let \( y = x \) be the equation of a chord of the circle \( C_1 \) (in the closed half-plane \( x \ge 0 \)) of diameter 10 passing through the origin. Let \( C_2 \) be another circle described on the given chord as diameter. If the equation of the chord of the circle \( C_2 \), which passes through the point \( (2, 3) \) and is farthest from the center of \( C_2 \), is \( x + ay + b = 0 \), then \( b \) is equal to:

• A. \( -2 \)
• B. \( 10 \)
• C. \( -6 \)
• D. \( 6 \)

Hint:

For maximum distance from the center, a chord passing through a fixed point must be perpendicular to the radius drawn to that point.

Answer 1

The Correct Option is D

Concept: For a circle, the chord farthest from the center and passing through a fixed point is perpendicular to the line joining the center to that point. Also, if a chord is taken as the diameter of another circle, the midpoint of the chord becomes the center of the new circle.
Step 1: Equation of the circle \( C_1 \) Diameter \( = 10 \Rightarrow \) radius \( = 5 \) Since the circle passes through the origin and is centered at the origin: \[ x^2 + y^2 = 25 \]
Step 2: Find the endpoints of the chord \( y = x \) Substitute \( y = x \) in the circle: \[ x^2 + x^2 = 25 \Rightarrow 2x^2 = 25 \Rightarrow x = \pm \frac{5}{\sqrt{2}} \] Thus, the chord has endpoints: \[ \left(\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}}\right),\; \left(-\frac{5}{\sqrt{2}}, -\frac{5}{\sqrt{2}}\right) \]
Step 3: Center of circle \( C_2 \) Since the chord is the diameter of \( C_2 \), its center is the midpoint of the endpoints: \[ \left(0, 0\right) \]
Step 4: Direction of the chord farthest from the center The chord of \( C_2 \) passing through \( (2, 3) \) and farthest from the center must be perpendicular to the line joining the center \( (0,0) \) to \( (2,3) \). Slope of line joining center to point: \[ \frac{3}{2} \] So, slope of the required chord: \[ -\frac{2}{3} \]
Step 5: Equation of the required chord Using point-slope form: \[ y - 3 = -\frac{2}{3}(x - 2) \] Simplifying: \[ 3y - 9 = -2x + 4 \] \[ 2x + 3y - 13 = 0 \] Dividing by 2 to match the given form: \[ x + \frac{3}{2}y - 6 = 0 \] Thus, \[ b = 6 \]