Question

A straight line \(y = x - 1\) intersects a circle with center at \(x = 1, y = 1\) and radius of magnitude 1 at two points. The length of the chord formed by this intersection is ________. (rounded off to three decimal places)

Hint:

Use \(2\sqrt{r^2-d^2}\), where \(d\) is the perpendicular distance from the center to the line, to get the chord length directly.

Answer 1

Correct Answer: 1.414

Step 1: Write the equation of the circle.
The circle has center \((1, 1)\) and radius \(1\), so its equation is
\[ (x-1)^2 + (y-1)^2 = 1 \]

Step 2: Substitute the line equation into the circle.
The line is \(y = x - 1\). Substituting:
\[ (x-1)^2 + (x - 1 - 1)^2 = 1 \]
\[ (x-1)^2 + (x-2)^2 = 1 \]

Step 3: Solve the resulting quadratic in \(x\).
Let \(u = x - 1\), so \(x - 2 = u - 1\). Then
\[ u^2 + (u-1)^2 = 1 \]
\[ u^2 + u^2 - 2u + 1 = 1 \]
\[ 2u^2 - 2u = 0 \]
\[ 2u(u - 1) = 0 \]
So \(u = 0\) or \(u = 1\), which means \(x = 1\) or \(x = 2\).

Step 4: Find the two intersection points.
Using \(y = x - 1\): at \(x = 1\), \(y = 0\), giving point \((1, 0)\); at \(x = 2\), \(y = 1\), giving point \((2, 1)\).

Step 5: Compute the chord length.
The chord length is the distance between \((1,0)\) and \((2,1)\):
\[ d = \sqrt{(2-1)^2 + (1-0)^2} = \sqrt{1 + 1} = \sqrt{2} \]

Final Answer:
\[ \sqrt{2} \approx 1.41421 \]
Rounded off to three decimal places:
\[ \boxed{1.414} \]