Maths · Equation of a circle when the endpoints of a diameter are given
Let \) and \) be two points. Let \) be a point such that 1) +(y-2)(y-4)=0 . \) I
Let \( \mathbf{A}(\mathbf{1}, \mathbf{2}) \) and \( \mathbf{B}(\mathbf{3}, \mathbf{4}) \) be two points. Let \( \mathbf{C}(x, y) \) be a point such that \( (x- \) 1) \( (x-3)+(y-2)(y-4)=0 . \) If area of \( (\Delta A B C)=1, \) then the maximum number of positions of \( \mathbf{C} \) in \( \boldsymbol{x} \boldsymbol{y} \) -plane is
- A. 2
- B. 4
- C. 8
- D. None of these
Step-by-step solution
The condition (x-1)(x-3)+(y-2)(y-4)=0 simplifies to the circle equation (x-2)^2+(y-3)^2=2, which has center at the midpoint of AB and radius √2. AB=2√2, so AB is a diameter. The area of triangle ABC is 1, so the distance from C to line AB must be 1/√2. The line AB is x-y+1=0, so the distance condition gives |x-y+1|=1. This yields two parallel lines: x-y=0 and x-y=-2. Each line intersects the circle at two distinct points (since the distance from the center to each line is 1/√2 < √2), resulting in 4 points total. Therefore, the maximum number of positions of C is 4.