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

Rather than solving for the two intersection points directly, this method uses the standard chord-length formula: if a line is at perpendicular distance $d$ from the center of a circle of radius $r$, the chord it cuts off has length $2\sqrt{r^2 - d^2}$.

First write the line $y = x - 1$ in general form:

$$ x - y - 1 = 0 $$

The center of the circle is $(1, 1)$ and the radius is $r = 1$. The perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is:

$$ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$

Here $A = 1$, $B = -1$, $C = -1$, and $(x_0, y_0) = (1, 1)$:

$$ d = \frac{|1(1) + (-1)(1) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1 - 1 - 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} $$

Now apply the chord-length formula:

$$ \text{chord length} = 2\sqrt{r^2 - d^2} = 2\sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^2} = 2\sqrt{1 - \frac{1}{2}} = 2\sqrt{\frac{1}{2}} $$

Simplify:

$$ 2\sqrt{\frac{1}{2}} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.41421 $$

Let's summarize:

• The perpendicular distance from the circle's center to the line is $1/\sqrt{2}$.
• Applying $2\sqrt{r^2 - d^2}$ with $r = 1$ gives the chord length directly as $\sqrt{2}$, without needing the actual coordinates of the intersection points.

So the chord length, rounded off to three decimal places, is $1.414$.