OpenSAT

If the circle intersects the line at exactly one point, the line is tangent to the circle. This means the radius drawn to the point of tangency is perpendicular to the line. The slope of the line y = 4x + b is 4, so the slope of the radius is –1/4. The radius passes through (0, 0), so the equation of the radius is y = –1/4x. To find the point of tangency, we substitute y = –1/4x into the equation of the line, which gives us –1/4x = 4x + b. Solving for x yields x = –b/17. Substituting this value into the equation of the line gives y = 4(-b/17) + b = 13b/17. The point of tangency has coordinates (-b/17, 13b/17). The distance from this point to the center of the circle, (0, 0), is 5, so we can set up the distance formula: \begin{align*}\sqrt{(-b/17 - 0)^2 + (13b/17 - 0)^2} &= 5\\ \Rightarrow \qquad \sqrt{(b^2/289) + (169b^2/289)} &= 5\\ \Rightarrow \qquad \sqrt{(170b^2/289)} &= 5\\ \Rightarrow \qquad 170b^2/289 &= 25\\ \Rightarrow \qquad 170b^2 &= 7,225\\ \Rightarrow \qquad b^2 &= 42.5\\ \Rightarrow \qquad b &= \pm 6.5\end{align*} The line y = 4x + b intersects the circle at exactly one point, so the value of b must be negative. Therefore, b = -6.5, which is equivalent to -20/3, or -20/3. Note that -20/3, -6.5, and -6.50 are all acceptable ways to enter a correct answer.