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

To solve the problem, we need to determine the image of a circle with diameter formed by the points of intersection of a line and another circle, and then find its image in a specific line. Let's break down the solution step-by-step:

1. Identify the points of intersection of the line \(a x + b y = 0\) with the circle \(x^2 + y^2 - 2x = 0\):
   • Rewrite the equation of the circle as \((x-1)^2 + y^2 = 1\), which is a circle with center \((1, 0)\) and radius \(1\).
   • The line can be written as \(y = -\frac{a}{b}x\) since \(a \neq b\).
   • Substitute \(y = -\frac{a}{b}x\) into the circle's equation:
2. Find points \(A(\alpha, 0)\) and \(B(1, \beta)\):
   • The equations \(\alpha = 0\) and AB:
     • The center is at \(\left(\frac{\alpha + 1}{2}, \frac{0 + \beta}{2}\right)\).
   • Find the image of the circle about the line \(x + y + 2 = 0\):
     • Use the formula for reflection of a point across a line. Let's find the image point of the center first.
     • For point \((x_0, y_0)\) reflected in the line \(ax + by + c = 0\), the image is given by the formula:
   • Determine center and radius of the image circle:
   • Use circle's standard form \((x - x')^2 + (y - y')^2 = r^2\).
   • Find the option that matches the circle's equation with \(x^2 + y^2 + 5x + 5y + 12 = 0\).

Therefore, the image of the required circle is

$x^2 + y^2 + 5x + 5y + 12 = 0$