Question

what is the center of the circle with the equation $(x + 6)^2+(y - 5)^2=100$?
a. $(6,5)$
b. $(-6,5)$
c. $(6,-5)$
d. $(-6,-5)$
which of the following is the equation of a circle with center $(0,-1)$ and radius 6?
a. $x^2+(y + 1)^2=12$
b. $x^2+y^2=12$
c. $x^2+y^2=36$
d. $x^2+(y + 1)^2=36$

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: Identify the center from the given equation \((x + 6)^2+(y - 5)^2 = 100\)

Compare with \((x - h)^2+(y - k)^2=r^2\). Here \(x+6=(x - (- 6))\) and \(y - 5\). So \(h=-6,k = 5\). The center is \((-6,5)\) (option b).

Step3: Recall the standard form for the second part

For a circle with center \((0,-1)\) and radius \(r = 6\), using \((x - h)^2+(y - k)^2=r^2\), we have \(h = 0,k=-1,r = 6\). Substitute into the formula: \((x-0)^2+(y-(-1))^2=6^2\), which simplifies to \(x^{2}+(y + 1)^2=36\) (option d).

Answer:

For the first question: b. \((-6,5)\)
For the second question: d. \(x^{2}+(y + 1)^2=36\)