Question

The equation of the smallest circle passing through the intersection of x2 + y2 – 2x – 4y – 4 = 0 and the line x + y – 4 = 0 is –

• A. x2 + y2 – 3x – 5y – 8 = 0
• B. x2 + y2 – x – 3y = 0
• C. x2 + y2 – 3x – 5y = 0
• D. x2 + y2 – x – 3y – 8 = 0

Answer 1

Answer: (C)

x2 + y2 – 3x – 5y = 0

Answer 2

The required circle must have the common chord as its diameter. The family of circles with x + y – 4 = 0 as common chord is x2 + y2 + x (l – 2) + y (l – 4) – (4l + 4) = 0

The centre of this circle, namely, $\left( \frac{2–\lambda}{2},\frac{4–\lambda}{2} \right)$ should lie on x + y – 4 = 0

\ 1 –$\frac{\lambda}{2}$+ 2 – $\frac{\lambda}{2}$ – 4 = 0

\ l = – 1 Required circle is x2 + y2 – 3x – 5y = 0.