Question

write the equation of the circle centered at (-3,5) with diameter 12. add work check answer

Explanation:

Step1: Recall the standard form of a circle's equation

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

Step2: Identify the center \((h,k)\)

Given the center \((-3,5)\), so \(h=-3\), \(k = 5\).

Step3: Calculate the radius \(r\)

Since the diameter \(d = 12\), and \(r=\frac{d}{2}\), then \(r=\frac{12}{2}=6\).

Step4: Substitute \(h,k,r\) into the standard form

Substitute \(h=-3\), \(k = 5\), \(r = 6\) into \((x - h)^2+(y - k)^2=r^2\).
We get \((x-(-3))^2+(y - 5)^2=6^2\), which simplifies to \((x + 3)^2+(y - 5)^2=36\).

Answer:

\((x + 3)^2+(y - 5)^2=36\)