Question

Consider a circle \(C_1\), passing through origin and lying in region \(0 \le x\) only, with diameter = 10. Consider a chord \(PQ\) of \(C_1\) with equation \(y = x\) and another Circle \(C_2\) which has \(PQ\) as diameter. A chord is drawn to \(C_2\) passing through (2, 3) such that distance of chord from centre of \(C_2\) is maximum has equation \(x + ay + b = 0\) then \(|b - a|\) is equal to :

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The chord of a circle that is at maximum distance from the center and passes through a point inside the circle is always perpendicular to the line joining that point and the center.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The chord with maximum distance from the center through a given point \( M \) is the one perpendicular to the radius joining the center to \( M \).
Step 2: Key Formula or Approach:
\( C_1 \) has diameter 10 and passes through (0,0) in \( x \ge 0 \). Its center is (5,0).
Equation of \( C_1 \): \( (x-5)^2 + y^2 = 25 \).
Find \( P, Q \) by intersecting \( C_1 \) with \( y = x \).
Step 3: Detailed Explanation:
\( (x-5)^2 + x^2 = 25 \implies 2x^2 - 10x = 0 \implies x=0, 5 \).
Points are \( P(0,0) \) and \( Q(5,5) \).
Center of \( C_2 \) is midpoint of \( PQ \), which is \( C(2.5, 2.5) \).
Chord of \( C_2 \) passes through \( M(2, 3) \).
Slope of \( CM = \frac{3 - 2.5}{2 - 2.5} = \frac{0.5}{-0.5} = -1 \).
Slope of required chord \( = 1 \) (perpendicular to \( CM \)).
Equation of chord: \( y - 3 = 1(x - 2) \implies x - y + 1 = 0 \).
Comparing with \( x + ay + b = 0 \): \( a = -1, b = 1 \).
\( |b - a| = |1 - (-1)| = 2 \).
Step 4: Final Answer:
\( |b - a| = 2 \).