Given circle E with diameter start overline, C, D, end overline
and radius start overline, E, A, end overline
. start overline, A, B, end overline
AB
is tangent to E at A. If E, C, equals, 11 and E, A, equals, 11, solve for A, CA. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."
A
D
C
E
B
Since AB is tangent to circle E at A, this means that the line segment AB is perpendicular to the radius EA drawn to the point of tangency A.
Therefore, triangle EAC is a right triangle with EA as the hypotenuse and AC as one of the legs. We can use the Pythagorean Theorem to find the length of AC.
AC^2 + EA^2 = EC^2
AC^2 + 11^2 = 11^2
AC^2 + 121 = 121
AC^2 = 0
AC = 0
Since the length of AC is 0, it means that A is the same point as C. Therefore, AC cannot be determined as it has no actual length.