The $x$-intercept is the value of $x$ when $y=0$, that is, the value of $x$ where the graph hits the $x$-axis.
\nTo find it, substitute $y=0$ into our equation:
\n\\[\\simplify[all]{(x-{a})^2}+\\simplify[basic]{(-{b})^2={r}^2}\\]
\nand solve for $x$:
\n| $\\simplify[all]{(x-{a})^2}$ | \n$=$ | \n$\\simplify[basic]{{r}^2-{abs(b)}^2}$ | \n
| $\\simplify[all]{(x-{a})}$ | \n$=$ | \n$\\pm\\sqrt{\\simplify[basic]{{r}^2-{abs(b)}^2}}$ | \n
| \n | $=$ | \n$\\pm\\sqrt{\\var{xdet}}$ | \n
| \n | $=$ | \n$\\pm$$\\var{xug}$ | \n
| $x$ | \n$=$ | \n$\\var{a}$$\\pm\\var{xug}$ | \n
| $x$ | \n$=$ | \n$\\var{a-xug}$$,\\var{a+xug}$ | \n
Since we are only working with real numbers, and the square root of a negative number is not a real number, there are no $x$-intercepts.
\nSince we took the (plus or minus) square root of zero we only have one $x$-intercept, $x=\\var{a}$.
\nIn this case we have two distinct $x$-intercepts.
\n\n
The $y$-intercept is the value of $y$ when $x=0$, that is, the value of $y$ where the graph hits the $y$-axis.
\nTo find it, substitute $x=0$ into our equation:
\n\\[\\simplify[basic]{(-{a})^2}+\\simplify[all]{(y-{b})^2}=\\simplify[basic]{{r}^2}\\]
\nand solve for $y$:
\n| $\\simplify[all]{(y-{b})^2}$ | \n$=$ | \n$\\simplify[basic]{{r}^2-{abs(a)}^2}$ | \n
| $\\simplify[all]{(y-{b})}$ | \n$=$ | \n$\\pm\\sqrt{\\simplify[basic]{{r}^2-{abs(a)}^2}}$ | \n
| \n | $=$ | \n$\\pm\\sqrt{\\var{ydet}}$ | \n
| \n | $=$ | \n$\\pm$$\\var{yug}$ | \n
| $y$ | \n$=$ | \n$\\var{b}$$\\pm\\var{yug}$ | \n
| $y$ | \n$=$ | \n$\\var{b-yug}$$,\\var{b+yug}$ | \n
Since we are only working with real numbers, and the square root of a negative number is not a real number, there are no $y$-intercepts.
\nSince we took the (plus or minus) square root of zero we only have one $y$-intercept, $y=\\var{b}$.
\nIn this case we have two distinct $y$-intercepts.
", "rulesets": {}, "parts": [{"prompt": "The set of $x$-intercepts of the graph of this equation would be [[0]].
\nNote: If there are no intercepts, enter set()
\nIf there is only one intercept, say $x=5$, enter set(5)
\nIf there are two intercepts, say $x=-2$ and $x=1.5$, enter set(-2,1.5)
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\nNote: If there are no intercepts, enter set()
\nIf there is only one intercept, say $y=5$, enter set(5)
\nIf there are two intercepts, say $y=-2$ and $y=1.5$, enter set(-2,1.5)
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