a radius 6 and center $(0,2)$ ? A small radio transmitter broadcasts in a 53 mile radius. If you drive along a straight line from a city 70 miles north of the transmitter to a second city 74 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Step 1 ：The problem can be solved using the Pythagorean theorem and the properties of a circle. The signal from the transmitter can be picked up within a circle with a radius of 53 miles. The cities are located outside this circle, and the straight line between them intersects the circle. The length of the drive during which the signal can be picked up is equal to the length of the line segment that is inside the circle.

Step 2 ：To find this length, we can first find the distance from the transmitter to the line between the cities. This is the shortest distance from the center of the circle to the line, and it can be found using the formula for the distance from a point to a line in the plane. If the line passes through the points \((x_1, y_1)\) and \((x_2, y_2)\), and the point is \((x_0, y_0)\), then the distance is \[d = \frac{|(y_2 - y_1)x_0 - (x_2 - x_1)y_0 + x_2y_1 - y_2x_1|}{\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}}.\]

Step 3 ：In this case, the line passes through the points \((0, 70)\) and \((74, 0)\), and the point is \((0, 0)\), so the distance is \[d = \frac{|70*74|}{\sqrt{70^2 + 74^2}}.\]

Step 4 ：If this distance is less than 53 miles, then the line intersects the circle, and the length of the line segment inside the circle is \[l = 2\sqrt{53^2 - d^2}.\] If the distance is greater than 53 miles, then the line does not intersect the circle, and the length of the line segment inside the circle is 0.

Step 5 ：By calculating, we get the radius \(r = 53\), the distance \(d = 50.85278918683011\), and the length \(l = 29.865955998080153\).

Step 6 ：The length of the drive during which you will pick up a signal from the transmitter is approximately \(\boxed{29.87}\) miles.