Question

If \( (1,1), (-2,2), (2,-2) \) are 3 points on a circle \( S \), then the perpendicular distance from the center of the circle \( S \) to the line \( 3x - 4y + 1 = 0 \) is:

• A. \( \frac{1}{2} \)
• B. \( 1 \)
• C. \( \frac{23}{10} \)
• D. \( 2 \)

Hint:

Use the determinant method to find the equation of a circle from three points, then use the perpendicular distance formula.

Answer 1

The Correct Option is A

Let the three points be \(A(1, 1)\), \(B(-2, 2)\), and \(C(2, -2)\).
We want to find the center of the circle passing through these points.
Let the center be \((h, k)\).
The distance from the center to each of the points is the radius \(r\).
So, \((h-1)^2 + (k-1)^2 = (h+2)^2 + (k-2)^2 = (h-2)^2 + (k+2)^2\).
From \((h-1)^2 + (k-1)^2 = (h+2)^2 + (k-2)^2\):
\(h^2 - 2h + 1 + k^2 - 2k + 1 = h^2 + 4h + 4 + k^2 - 4k + 4\)
\(-2h - 2k + 2 = 4h - 4k + 8\)
\(6h - 2k + 6 = 0\)
\(3h - k + 3 = 0\)
\(k = 3h + 3 \quad (1)\)
From \((h+2)^2 + (k-2)^2 = (h-2)^2 + (k+2)^2\):
\(h^2 + 4h + 4 + k^2 - 4k + 4 = h^2 - 4h + 4 + k^2 + 4k + 4\)
\(4h - 4k + 8 = -4h + 4k + 8\)
\(8h - 8k = 0\)
\(h = k \quad (2)\)
Substituting (2) into (1):
\(h = 3h + 3\)
\(-2h = 3\)
\(h = -\frac{3}{2}\)
Since \(h = k\), \(k = -\frac{3}{2}\).
The center of the circle is \(\left(-\frac{3}{2}, -\frac{3}{2}\right)\).
We want to find the perpendicular distance from the center \(\left(-\frac{3}{2}, -\frac{3}{2}\right)\) to the line \(3x - 4y + 1 = 0\).
The distance is given by:
\[d = \frac{|3(-\frac{3}{2}) - 4(-\frac{3}{2}) + 1|}{\sqrt{3^2 + (-4)^2}}\] \[d = \frac{|-\frac{9}{2} + \frac{12}{2} + 1|}{\sqrt{9 + 16}}\] \[d = \frac{|\frac{3}{2} + 1|}{5}\] \[d = \frac{|\frac{5}{2}|}{5}\] \[d = \frac{5}{2 \times 5} = \frac{1}{2}\] Final Answer: The final answer is $\boxed{(1)}$