QUESTION IMAGE
Question
give the center and radius of the circle described by the equation and graph the equation. use the graph to identify the domain and range.
$(x + 6)^{2}+(y - 5)^{2}=25$
the center is $(-6,5)$
(type an ordered pair. simplify your answer.)
the radius is 5
(type an integer or a simplified fraction.)
graph the circle
Step1: Recall the standard form of a circle 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
For the equation \((x + 6)^2+(y - 5)^2=25\), we can rewrite \((x + 6)\) as \((x-(-6))\). Comparing with \((x - h)^2+(y - k)^2=r^2\), we get \(h=-6\) and \(k = 5\). So the center \((h,k)=(-6,5)\).
Step3: Identify the radius
Since \(r^2 = 25\), then \(r=\sqrt{25}=5\) (we take the positive value as radius is non - negative).
Step4: Find the domain
The domain of a circle \((x - h)^2+(y - k)^2=r^2\) is given by \(h - r\leqslant x\leqslant h + r\). Substituting \(h=-6\) and \(r = 5\), we have \(-6-5\leqslant x\leqslant-6 + 5\), which simplifies to \(-11\leqslant x\leqslant-1\).
Step5: Find the range
The range of a circle \((x - h)^2+(y - k)^2=r^2\) is given by \(k - r\leqslant y\leqslant k + r\). Substituting \(k = 5\) and \(r = 5\), we have \(5-5\leqslant y\leqslant5 + 5\), which simplifies to \(0\leqslant y\leqslant10\).
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- Center: \((-6,5)\)
- Radius: \(5\)
- Domain: \([-11,-1]\)
- Range: \([0,10]\)