Question

The equation of circle with centre (2, -3) and touching x-axis is

• A. \( x^2+y^2-4x-6y+4=0 \)
• B. \( x^2+y^2-4x-6y-8=0 \)
• C. \( x^2+y^2-4x+6y+4=0 \)
• D. \( x^2+y^2+4x-6y+8=0 \)

Hint:

Equation of circle with center (h,k) and radius r: \((x-h)^2+(y-k)^2=r^2\).
If a circle touches the x-axis, its radius is \(|k|\) (absolute value of y-coordinate of center).
If a circle touches the y-axis, its radius is \(|h|\) (absolute value of x-coordinate of center).

Answer 1

The Correct Option is C

The center of the circle is \((h,k) = (2, -3)\). The circle touches the x-axis. The x-axis is the line \(y=0\). The distance from the center \((h,k)\) to the x-axis is \(|k|\). Since the circle touches the x-axis, its radius \(r\) must be equal to the perpendicular distance from its center to the x-axis. So, \(r = |k| = |-3| = 3\). The equation of a circle with center \((h,k)\) and radius \(r\) is \((x-h)^2 + (y-k)^2 = r^2\). Substitute \(h=2, k=-3, r=3\): \((x-2)^2 + (y-(-3))^2 = 3^2\) \((x-2)^2 + (y+3)^2 = 9\) Expand the equation: \( (x^2 - 4x + 4) + (y^2 + 6y + 9) = 9 \) \( x^2 - 4x + 4 + y^2 + 6y + 9 - 9 = 0 \) \( x^2 + y^2 - 4x + 6y + 4 = 0 \) This matches option (c). \[ \boxed{x^2+y^2-4x+6y+4=0} \]