Question

The points of intersection of the line $a x+b y=0,(a \neq b)$ and the circle $x^2+y^2-2 x=0$ are $A (\alpha, 0)$ and $B (1, \beta)$ The image of the circle with $AB$ as a diameter in the line $x+y+2=0$ is :

• A. $x^2+y^2+3 x+3 y+4=0$
• B. $x^2+y^2+5 x+5 y+12=0$
• C. $x^2+y^2+3 x+5 y+8=0$
• D. $x^2+y^2-5 x-5 y+12=0$

Answer 1

The Correct Option is B

\(\text{Only possibility}\) \(α=0,β=1 \)
\(∴\text{ equation of circle}\; x^2+y^2−x−y=0\) 
\(\text{Image of circle in}\; x+y+2=0 \;is \)
\(x^2+y^2+5x+5y+12=0\)

\(\text{Hence, the correct option is(B):}\) \(x^2+y^2+5x+5y+12=0\)

Answer 2

1. The given circle is: \[ x^2 + y^2 - 2x = 0. \] Rewriting it in the standard form: \[ (x - 1)^2 + y^2 = 1. \] The center is \((1, 0)\), and the radius is \(1\). 2. The line \(ax + by = 0\) intersects the circle at points \(A\) and \(B\). Using the parametric form of the circle, we find \(A\) and \(B\). Here, \(AB\) is the diameter. 3. The circle formed by the reflection of this circle in the line \(x + y + 2 = 0\) will have its center reflected. The reflection of the center \((1, 0)\) in the line \(x + y + 2 = 0\) is calculated using the formula for reflection. 4. The resulting equation for the new circle is: \[ x^2 + y^2 + 5x + 5y + 12 = 0. \] Thus, the correct answer is: \[ x^2 + y^2 + 5x + 5y + 12 = 0. \] The problem combines geometry of circles and reflections. The reflection of the center is used to derive the equation of the image circle.