Given: KL is a tangent to the circle. LM is another tangent to the circle.
We can use the tangent meeting at an external point theorem. <em>Theorem</em>: <em>The tangent segments to a circle from an external point are equal. </em> Thus, we can say KL = LM, as they lie on a common circle, and are tangents to such circle.
4x - 2 = 3x + 3 4x - 3x = 3 + 2 x = 5
Since LM is 3x + 3, we can substitute the value of x to LM to render: 3(5) + 3 = 18