Question

question
determine the equation of the circle with center (0, 0) containing the point (5, √24).

Explanation:

Step1: Recall the circle equation formula

The standard equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). For a circle with center \((0,0)\), the equation simplifies to \(x^{2}+y^{2}=r^{2}\), where \(r\) is the radius.

Step2: Calculate the radius

The radius \(r\) is the distance from the center \((0,0)\) to the point \((5,\sqrt{24})\) on the circle. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), with \((x_1,y_1)=(0,0)\) and \((x_2,y_2)=(5,\sqrt{24})\), we have:
\[

$$\begin{align*} r&=\sqrt{(5 - 0)^2+(\sqrt{24}-0)^2}\\ &=\sqrt{5^{2}+(\sqrt{24})^{2}}\\ &=\sqrt{25 + 24}\\ &=\sqrt{49}\\ & = 7 \end{align*}$$

\]

Step3: Substitute \(r\) into the circle equation

Since \(r = 7\) and the center is \((0,0)\), substituting into \(x^{2}+y^{2}=r^{2}\) gives:
\[x^{2}+y^{2}=7^{2}=49\]

Answer:

The equation of the circle is \(x^{2}+y^{2}=49\)