Step 1: Understanding the Concept:
We are tasked with finding an unknown point that lies on a specific circle.
To do this, we first need to determine the full equation of the circle using the geometric constraints provided: a point it passes through and a point where it is tangent to an axis.
Step 2: Key Formula or Approach:
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.
If a circle is tangent to the x-axis at a point \((x_1, 0)\), its center lies directly above or below this point, so \(h = x_1\).
Furthermore, the distance from the center to the tangent x-axis is the radius, meaning \(r = |k|\), or \(r^2 = k^2\).
Step 3: Detailed Explanation:
The problem states the circle touches the x-axis at the point \((3, 0)\).
This geometric fact gives us two pieces of information:
1. The x-coordinate of the center \((h, k)\) must be \(h = 3\).
2. The radius \(r\) is the perpendicular distance to the x-axis, so \(r = |k|\), which implies \(r^2 = k^2\).
Incorporating these into the standard circle equation, we get a simplified form:
\[ (x - 3)^2 + (y - k)^2 = k^2 \]
We are also given that the circle passes through the point \((1, -2)\). This point must satisfy the circle's equation.
Substitute \(x = 1\) and \(y = -2\) into our simplified equation:
\[ (1 - 3)^2 + (-2 - k)^2 = k^2 \]
Simplify the terms inside the parentheses:
\[ (-2)^2 + (-(2 + k))^2 = k^2 \]
Since \((-(2+k))^2\) is equivalent to \((2+k)^2\), we expand the squares:
\[ 4 + (4 + 4k + k^2) = k^2 \]
Combine the constant terms:
\[ 8 + 4k + k^2 = k^2 \]
Subtract \(k^2\) from both sides of the equation to solve for \(k\):
\[ 8 + 4k = 0 \]
\[ 4k = -8 \]
\[ k = -2 \]
Now we know the center is \((3, -2)\) and the radius squared is \(r^2 = (-2)^2 = 4\).
The complete equation of the circle is therefore:
\[ (x - 3)^2 + (y + 2)^2 = 4 \]
Finally, we must check which of the given options satisfies this equation.
Testing Option (A) \((2, -5)\): \((2 - 3)^2 + (-5 + 2)^2 = (-1)^2 + (-3)^2 = 1 + 9 = 10 \neq 4\).
Testing Option (B) \((-5, -2)\): \((-5 - 3)^2 + (-2 + 2)^2 = (-8)^2 + 0^2 = 64 \neq 4\).
Testing Option (C) \((-2, 5)\): \((-2 - 3)^2 + (5 + 2)^2 = (-5)^2 + 7^2 = 25 + 49 = 74 \neq 4\).
Testing Option (D) \((-5, 2)\): \((-5 - 3)^2 + (2 + 2)^2 = (-8)^2 + 4^2 = 64 + 16 = 80 \neq 4\).
Testing Option (E) \((5, -2)\): \((5 - 3)^2 + (-2 + 2)^2 = (2)^2 + (0)^2 = 4 + 0 = 4\).
Since \(4 = 4\), the point \((5, -2)\) perfectly satisfies the equation of the circle.
Step 4: Final Answer:
The circle also passes through the point \((5, -2)\).