Question

1. o is the centre of this circle and point a is a point of tangency. determine the value of b. if necessary, give your answer to the nearest tenth.

circle with center o, point a on the circle, tangent from b to a, oa = b, ob = 30, ab = 19

Explanation:

Step1: Recall tangent - radius property

A tangent to a circle is perpendicular to the radius at the point of tangency. So, \(\angle OAB = 90^{\circ}\), which means triangle \(OAB\) is a right - triangle with right angle at \(A\).

Step2: Apply Pythagorean theorem

In right - triangle \(OAB\), we know that the Pythagorean theorem states that for a right - triangle with hypotenuse \(c\) and legs \(a\) and \(b\), \(c^{2}=a^{2}+b^{2}\). Here, \(OB = 30\), \(AB = 19\), and \(OA=b\) (the radius), \(OB\) is the hypotenuse. So, we can rewrite the Pythagorean theorem as \(b^{2}+19^{2}=30^{2}\).
First, calculate \(19^{2}=361\) and \(30^{2} = 900\). Then, \(b^{2}=30^{2}-19^{2}=900 - 361=539\).

Step3: Solve for \(b\)

Take the square root of both sides: \(b=\sqrt{539}\approx23.2\) (rounded to the nearest tenth).

Answer:

\(b\approx23.2\)