Question

in the figure below, the segments \\(\overline{ij}\\) and \\(\overline{ik}\\) are tangent to the circle centered at \\(o\\). given that \\(oj = 2\\) and \\(ik = 2.1\\), find \\(oi\\).

Explanation:

Step1: Recall tangent properties

Tangents from a common external point to a circle are equal. So, \( IJ = IK = 2.1 \). Also, the radius is perpendicular to the tangent at the point of contact, so \( \angle OJI = 90^\circ \), making \( \triangle OJI \) a right triangle.

Step2: Apply Pythagorean theorem

In right triangle \( \triangle OJI \), \( OJ = 2 \) (radius), \( IJ = 2.1 \) (tangent), and \( OI \) is the hypotenuse. By Pythagorean theorem, \( OI=\sqrt{OJ^{2}+IJ^{2}} \).
Substitute \( OJ = 2 \) and \( IJ = 2.1 \):
\( OI=\sqrt{2^{2}+2.1^{2}}=\sqrt{4 + 4.41}=\sqrt{8.41} \)

Step3: Calculate the square root

\( \sqrt{8.41}=2.9 \)

Answer:

\( 2.9 \)