From the point, the shortest distance of which to the circle is 25 mm, draw a tangent to the circle.

From the point, the shortest distance of which to the circle is 25 mm, draw a tangent to the circle. The segment of this tangent between this point and the point of tangency is 35 mm. find the length of the diameter of the circle.

Construct the radius OB, which is perpendicular to the tangent AB by the property of the tangent to the circle. Then the triangle ABO is rectangular.

Segment OA = OC = R.

Then, the segment AO = AC + OC = 25 + R.

Then, by the Pythagorean theorem, AB ^ 2 = AO ^ 2 + OB ^ 2 = (25 + R) ^ 2 + R ^ 2 = 35 ^ 2.

1225 = 625 + 50 * R + R ^ 2 + R ^ 2.

2 * R ^ 2 + 50 * R – 600 = 0.

R ^ 2 + 25 * R – 300 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = 25 ^ 2 – 4 * 1 * (-600) = 625 + 2400 = 3025.

X1 = (-25 – √3025) / (2 * 1) = (-25 – 55) / 2 = -80 / 2 = -40. (Doesn’t fit because <0).

X2 = (-25 + √3025) / (2 * 1) = (-25 + 55) / 2 = 30/2 = 15.

R = X2 = 15 cm.

Answer: The radius of the circle is 15 cm.