Question

1. if the point $(-5,b\sqrt{3})$ lies on a circle in quadrant ii with a radius of 10 units, determine a possible value(s) of $b$. the value of $b$ to the nearest tenth, is:________

Explanation:

Step1: Use the equation of a circle

The equation of a circle with center \((0,0)\) is \(x^{2}+y^{2}=r^{2}\). Here \(x = - 5\), \(y=b\sqrt{3}\), and \(r = 10\). Substitute these values into the equation: \((-5)^{2}+(b\sqrt{3})^{2}=10^{2}\).

Step2: Simplify the equation

Expand the terms: \(25 + 3b^{2}=100\).

Step3: Solve for \(b^{2}\)

Subtract \(25\) from both sides: \(3b^{2}=100 - 25=75\). Then \(b^{2}=\frac{75}{3}=25\).

Step4: Solve for \(b\)

Take the square - root of both sides: \(b=\pm5\). Since the point \((-5,b\sqrt{3})\) is in quadrant II, \(y=b\sqrt{3}>0\). If \(b\sqrt{3}>0\) and \(\sqrt{3}>0\), then \(b>0\). So \(b = 5.0\).

Answer:

\(5.0\)