Question

The equation of the circle which touches the x-axis, passes through the point (1, 1) and whose centre lies on the line $ x + y = 3 $ in the first quadrant is

• A. \( x^2 + y^2 + 4x + 2y + 4 = 0 \)
• B. \( x^2 + y^2 - 4x - 2y + 4 = 0 \)
• C. \( x^2 + y^2 + 4x - 2y + 4 = 0 \)
• D. \( x^2 + y^2 - 4x + 2y + 4 = 0 \)

Hint:

When a circle touches the x-axis, its radius is equal to the y-coordinate of its center. Use the point on the circle and the equation of the line to find the center and the radius.

Answer 1

The Correct Option is B

The general equation of a circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \( (h, k) \) is the center of the circle and \( r \) is its radius. Given that the circle touches the x-axis, the radius of the circle is equal to the y-coordinate of its center, i.e., \( k = r \). Since the center lies on the line \( x + y = 3 \), we have: \[ h + k = 3 \quad \text{or} \quad h = 3 - k. \] We are also given that the circle passes through the point \( (1, 1) \). Substituting this point into the equation of the circle, we get: \[ (1 - h)^2 + (1 - k)^2 = r^2. \] Now, substitute \( h = 3 - k \) and \( r = k \) into the above equation: \[ (1 - (3 - k))^2 + (1 - k)^2 = k^2. \] Simplifying: \[ (1 - 3 + k)^2 + (1 - k)^2 = k^2 \quad \Rightarrow \quad (k - 2)^2 + (1 - k)^2 = k^2. \] Expanding: \[ (k^2 - 4k + 4) + (1 - 2k + k^2) = k^2 \quad \Rightarrow \quad 2k^2 - 6k + 5 = k^2. \] Solving for \( k \): \[ k^2 - 6k + 5 = 0. \] Using the quadratic formula: \[ k = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}. \] So, \( k = 5 \) or \( k = 1 \). Since the circle is in the first quadrant, we take \( k = 1 \), and therefore \( h = 3 - k = 2 \). Now, the equation of the circle is: \[ (x - 2)^2 + (y - 1)^2 = 1^2 \quad \Rightarrow \quad (x - 2)^2 + (y - 1)^2 = 1. \] Expanding: \[ (x^2 - 4x + 4) + (y^2 - 2y + 1) = 1 \quad \Rightarrow \quad x^2 + y^2 - 4x - 2y + 5 = 1 \quad \Rightarrow \quad x^2 + y^2 - 4x - 2y + 4 = 0. \] Thus, the correct equation is: \[ x^2 + y^2 - 4x - 2y + 4 = 0. \] Hence, the correct answer is (B).