Digital SAT Math Practice Question #1180

⚡ **Geometric Shortcut:** Instead of finding the intersection points, use the relationship between the circle's radius, the distance from the center to the chord, and half the chord's length. 1. **Radius (r):** Given as 5. 2. **Distance (d) from center to line:** The center is the origin (0,0). The line is \( y = x + 1 \), which can be written as \( x - y + 1 = 0 \). The distance formula from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is \( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \). \( d = \frac{|1(0) - 1(0) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1|}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}} \). 3. **Pythagorean Theorem:** The radius, the distance \( d \), and half the chord length (\( L/2 \)) form a right-angled triangle, where the radius is the hypotenuse. So, \( r^2 = d^2 + (L/2)^2 \). \( 5^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{L}{2}\right)^2 \) \( 25 = \frac{1}{2} + \frac{L^2}{4} \) 4. **Solve for L:** Multiply the entire equation by 4 to eliminate denominators: \( 100 = 2 + L^2 \) \( L^2 = 98 \) \( L = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2} \). This method often saves time by avoiding quadratic equations and potentially complex coordinates.