Question

the figure shows a circle of radius 2 along with four labeled points in the xy-plane. the measure of angle cob is equal to the measure of angle aob. what are the coordinates of point b?

Explanation:

Step1: Determine the angle of OA

Point A is at (0, -2), so the angle of OA with the positive x - axis is \(270^{\circ}\) or \(\frac{3\pi}{2}\) radians.

Step2: Analyze the angle relationship

Given that \(m\angle COB=m\angle AOB\). The angle of OC with the positive x - axis is \(0^{\circ}\) (since C is at (2, 0)). Let the angle of OB with the positive x - axis be \(\theta\). The angle between OA and OB is \(\frac{3\pi}{2}-\theta\), and the angle between OB and OC is \(\theta - 0=\theta\). Since \(m\angle COB = m\angle AOB\), we have \(\theta=\frac{3\pi}{2}-\theta\). Solving for \(\theta\):
\[ \begin{align*} \theta+\theta&=\frac{3\pi}{2}\\ 2\theta&=\frac{3\pi}{2}\\ \theta&=\frac{3\pi}{4} \end{align*} \]

Step3: Use polar - rectangular conversion

For a point on a circle of radius \(r = 2\) with angle \(\theta=\frac{3\pi}{4}\) from the positive x - axis, the rectangular coordinates \((x,y)\) are given by \(x = r\cos\theta\) and \(y=r\sin\theta\).
- Calculate \(x\): \(x = 2\cos\frac{3\pi}{4}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\)? Wait, no, wait. Wait, we made a mistake in the angle calculation. Wait, point A is at (0, - 2), so the angle of OA is \(270^{\circ}\) (or \(\frac{3\pi}{2}\)). The angle between OA and the negative y - axis is \(0\), and the angle between OB and OA is equal to the angle between OB and OC. Let's re - consider. The angle of OC is \(0^{\circ}\), angle of OA is \(270^{\circ}\). The total angle between OC and OA is \(270^{\circ}\). Since \(\angle COB=\angle AOB\), each angle is \(\frac{270^{\circ}}{2} = 135^{\circ}\) from OC? Wait, no. Wait, the angle between OC (along positive x - axis) and OB, and between OB and OA (along negative y - axis) are equal. So the angle of OB with respect to the positive x - axis is \(360^{\circ}- 135^{\circ}=225^{\circ}\)? No, wait, let's use symmetry. The line OA is along the negative y - axis, OC is along positive x - axis. The angle between OC and OA is \(90^{\circ}+90^{\circ}=270^{\circ}\)? No, the angle between positive x - axis (OC) and negative y - axis (OA) is \(270^{\circ}\) or \(\frac{3\pi}{2}\) radians. If \(\angle COB=\angle AOB\), then the angle of OB is \(\frac{3\pi}{2}-\frac{3\pi}{4}=\frac{3\pi}{4}\)? No, wait, let's use the unit circle concept. The radius of the circle is \(r = 2\). Point A is (0, - 2), so the vector OA is \((0,-2)\), point C is (2, 0), vector OC is \((2,0)\). Let point B be \((x,y)\) on the circle \(x^{2}+y^{2}=4\) (since radius is 2). The angle between OC and OB is equal to the angle between OB and OA. The angle between two vectors \(\vec{u}=(x_1,y_1)\) and \(\vec{v}=(x_2,y_2)\) is given by \(\cos\alpha=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}\). Let \(\vec{OC}=(2,0)\), \(\vec{OB}=(x,y)\), \(\vec{OA}=(0, - 2)\). Then \(\cos\angle COB=\frac{\vec{OC}\cdot\vec{OB}}{\vert\vec{OC}\vert\vert\vec{OB}\vert}=\frac{2x}{2\times2}=\frac{x}{2}\), and \(\cos\angle AOB=\frac{\vec{OB}\cdot\vec{OA}}{\vert\vec{OB}\vert\vert\vec{OA}\vert}=\frac{- 2y}{2\times2}=\frac{-y}{2}\). Since \(\angle COB=\angle AOB\), \(\cos\angle COB=\cos\angle AOB\), so \(\frac{x}{2}=\frac{-y}{2}\), which implies \(x=-y\). Also, since \(x^{2}+y^{2}=4\) (because B is on the circle of radius 2 centered at the origin), substitute \(y=-x\) into the equation:
\[ \begin{align*} x^{2}+(-x)^{2}&=4\\ 2x^{2}&=4\\ x^{2}&=2\\ x&=\sqrt{2}\text{ or }x = -\sqrt{2} \end{align*} \]
Since point B is in the fourth quadrant (from the figure), \(x>0\) and \(y < 0\). If \(x=\sqrt{2}\), then \(y=-x=-\sqrt{2}\)? Wait, no, wait. Wait, if \(x=-y\), and B is in the fourth quadrant, x is positive, y is negative. So if \(x = \sqrt{2}\), then \(y=-\sqrt{2}\)? Wait, but let's check the angle. The angle of \((\sqrt{2},-\sqrt{2})\) with the positive x - axis is \(-45^{\circ}\) or \(315^{\circ}\). The angle between OC (0 degrees) and OB (315 degrees) is \(45^{\circ}\), and the angle between OB (315 degrees) and OA (270 degrees) is also \(45^{\circ}\) (since \(315 - 270=45\)). Wait, we made a mistake earlier in the angle calculation. The angle between OC (0 degrees) and OA (270 degrees) is \(270\) degrees, but if \(\angle COB=\angle AOB\), then each angle is \(135\) degrees? No, that's wrong. Wait, the correct way is: the angle between OC (positive x - axis) and OB is \(\alpha\), and the angle between OB and OA (negative y - axis) is also \(\alpha\). So the sum of these two angles is the angle between OC and OA, which is \(90^{\circ}\) (from positive x - axis to negative y - axis is \(270^{\circ}\)? No, from positive x - axis to negative y - axis is \(270^{\circ}\), but if we consider the smaller angle, it's \(90^{\circ}\)? No, no. The positive x - axis to negative y - axis is \(270^{\circ}\) in the counter - clockwise direction or \(90^{\circ}\) in the clockwise direction. Wait, the figure shows that point B is in the fourth quadrant, below the x - axis and to the right of the y - axis. So the angle of OB with respect to the positive x - axis is \(360^{\circ}- 45^{\circ}=315^{\circ}\), so the coordinates are \(x = 2\cos315^{\circ}=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\), \(y = 2\sin315^{\circ}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}\). Wait, but let's check the angle between OC and OB: the angle between (2,0) and \((\sqrt{2},-\sqrt{2})\) is calculated by \(\cos\theta=\frac{2\times\sqrt{2}+0\times(-\sqrt{2})}{2\times2}=\frac{\sqrt{2}}{2}\), so \(\theta = 45^{\circ}\). The angle between OB \((\sqrt{2},-\sqrt{2})\) and OA (0, - 2) is \(\cos\phi=\frac{0\times\sqrt{2}+(-2)\times(-\sqrt{2})}{2\times2}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\), so \(\phi = 45^{\circ}\). So \(\angle COB=\angle AOB = 45^{\circ}\), which satisfies the condition. So the coordinates of B are \((\sqrt{2},-\sqrt{2})\).

Answer:

\((\sqrt{2},-\sqrt{2})\)