Question

A circle touches the y-axis at (0, 4) and passes through the point (-2, 0). Then the radius of the circle is

• A. 5
• B. 2.4
• C. 3.7
• D. 4.6

Hint:

Visualizing the problem can be very helpful. Since the circle touches the y-axis at (0,4) and passes through (-2,0), the center must be in the second quadrant (negative h, positive k), which immediately tells you that \(h<0\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem can be solved using the standard equation of a circle and geometric properties of tangency.
Step 2: Key Formula or Approach:
1. The standard equation of a circle is \( (x-h)^2 + (y-k)^2 = r^2 \), where (h,k) is the center and r is the radius. 2. If a circle touches a vertical line (like the y-axis, x=0), the radius is equal to the absolute difference between the x-coordinate of the center and the line's x-value. 3. The radius to the point of tangency is perpendicular to the tangent line.
Step 3: Detailed Explanation:
Let the center of the circle be (h, k) and the radius be r.
The circle touches the y-axis (the line x=0) at the point (0, 4).
Since the tangent line (y-axis) is vertical, the radius to the point of tangency (0, 4) must be horizontal. This means the y-coordinate of the center must be the same as the y-coordinate of the point of tangency. Therefore, \( k = 4 \).
The distance from the center (h, 4) to the tangent line x=0 is the radius r. This distance is given by \( r = |h-0| = |h| \). So, the equation of the circle becomes: \[ (x-h)^2 + (y-4)^2 = r^2 = |h|^2 = h^2 \] We are given that the circle passes through the point (-2, 0). We can substitute these coordinates into the equation to find h. \[ (-2-h)^2 + (0-4)^2 = h^2 \] \[ (-(2+h))^2 + (-4)^2 = h^2 \] \[ (2+h)^2 + 16 = h^2 \] \[ (4 + 4h + h^2) + 16 = h^2 \] Now, we solve for h: \[ 4h + 20 + h^2 = h^2 \] \[ 4h + 20 = 0 \] \[ 4h = -20 \] \[ h = -5 \] The radius of the circle is \( r = |h| \). \[ r = |-5| = 5 \] Step 4: Final Answer:
The radius of the circle is 5.