2019 AMC 10 B

First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is , the Pythagorean Theorem gives .

Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center,, , and is cyclic. Therefore, we can apply Ptolemy's Theorem to give , where is the distance between the circle's center and . Therefore, . Using the Pythagorean Theorem on the triangle formed by the point , either one of or , and the circle's center, we find that , so , and thus the answer is .

We firstly obtain as in Solution . Label the point as . The midpoint of segment is . Notice that the center of the circle must lie on the line passing through the points and . Thus, the center of the circle lies on the line . Line is . Therefore, the slope of the line perpendicular to is , so its equation is .

But notice that this line must pass through and .

Hence . So the center of the circle is .

Finally, the distance between the center, , and point is . Thus the area of the circle is .

The midpoint of is . Let the tangent lines at and intersect at on the axis. Then is the perpendicular bisector of . . Let the center of the circle be . Then is similar to , so . The slope of is , so the slope of is . Hence, the equation of is . Letting , we have , so ..

Now, we compute , , and .

Therefore , and consequently, the area of the circle is .

Firstly, the point of intersection of the two tangent lines has an equal distance to points and due to power of a point theorem. This means we can easily find the point, which is . Label this point . is an isosceles triangle with lengths, , , and . Label the midpoint of segment as . The height of this triangle, or , is .

Since bisects , contains the diameter of circle . Let the two points on circle where intersects be and with being the shorter of the two. Now let be and be . By Power of a Point on and , . Applying Power of a Point again on and , . Expanding while using the fact that , . Plugging this into , . Using the quadratic formula, , and since , . Since this is the diameter, the radius of circle is , and so the area of circle is frac {85}{8}\pi$$.