Question

what is the radius of the circle with the equation $x^{2}+y^{2}=49$?
what is the radius of the circle with the equation $(x + 3)^{2}+(y - 2)^{2}=36$?

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: For the equation \(x^{2}+y^{2}=49\)

Rewrite it as \((x - 0)^2+(y - 0)^2 = 49\). Comparing with the standard form \((x - h)^2+(y - k)^2=r^2\), we have \(r^2=49\). Then \(r=\sqrt{49}=7\).

Step3: For the equation \((x + 3)^2+(y - 2)^2=36\)

Comparing with the standard form \((x - h)^2+(y - k)^2=r^2\), we have \(r^2 = 36\). Then \(r=\sqrt{36}=6\).

Answer:

For \(x^{2}+y^{2}=49\), the answer is a. 7. For \((x + 3)^2+(y - 2)^2=36\), the answer is b. 6.