{
"query": {
"display": "$$x^{2}-4x+y^{2}+6y-12=0$$",
"symbolab_question": "CONIC#x^{2}-4x+y^{2}+6y-12=0"
},
"solution": {
"level": "PERFORMED",
"subject": "Geometry",
"topic": "Circle",
"subTopic": "formula",
"default": "(a,b)=(2,-3),r=5"
},
"steps": {
"type": "interim",
"title": "$$x^{2}-4x+y^{2}+6y-12=0:\\quad$$Circle with center at $$\\left(2,\\:-3\\right)\\:$$and radius $$r=5$$",
"input": "x^{2}-4x+y^{2}+6y-12=0",
"steps": [
{
"type": "definition",
"title": "Circle Equation",
"text": "$$\\left(x−a\\right)^2+\\left(y−b\\right)^2=r^2\\:\\:$$is the circle equation with a radius r, centered at $$\\left(a,\\:b\\right)$$"
},
{
"type": "interim",
"title": "Rewrite $$x^{2}-4x+y^{2}+6y-12=0\\:$$in the form of the standard circle equation",
"input": "x^{2}-4x+y^{2}+6y-12=0",
"steps": [
{
"type": "step",
"primary": "Move the loose number to the right side",
"result": "x^{2}-4x+y^{2}+6y=12"
},
{
"type": "step",
"primary": "Group x-variables and y-variables together",
"result": "\\left(x^{2}-4x\\right)+\\left(y^{2}+6y\\right)=12"
},
{
"type": "step",
"primary": "Convert $$x\\:$$to square form",
"result": "\\left(x^{2}-4x+4\\right)+\\left(y^{2}+6y\\right)=12+4"
},
{
"type": "step",
"primary": "Convert to square form",
"result": "\\left(x-2\\right)^{2}+\\left(y^{2}+6y\\right)=12+4"
},
{
"type": "step",
"primary": "Convert $$y\\:$$to square form",
"result": "\\left(x-2\\right)^{2}+\\left(y^{2}+6y+9\\right)=12+4+9"
},
{
"type": "step",
"primary": "Convert to square form",
"result": "\\left(x-2\\right)^{2}+\\left(y+3\\right)^{2}=12+4+9"
},
{
"type": "step",
"primary": "Refine $$12+4+9$$",
"result": "\\left(x-2\\right)^{2}+\\left(y+3\\right)^{2}=25"
},
{
"type": "step",
"primary": "Rewrite in standard form",
"result": "\\left(x-2\\right)^{2}+\\left(y-\\left(-3\\right)\\right)^{2}=5^{2}"
}
],
"meta": {
"interimType": "Circle Canonical Format 1Eq"
}
},
{
"type": "step",
"result": "\\left(x-2\\right)^{2}+\\left(y-\\left(-3\\right)\\right)^{2}=5^{2}"
},
{
"type": "step",
"primary": "Therefore the circle properties are:",
"result": "\\left(a,\\:b\\right)=\\left(2,\\:-3\\right),\\:r=5"
}
],
"meta": {
"solvingClass": "Circle"
}
},
"plot_output": {
"meta": {
"plotInfo": {
"variable": "x",
"funcsToDraw": {
"funcs": [
{
"evalFormula": "y=\\sqrt{5^{2}-(x-2)^{2}}-3",
"displayFormula": "(x-2)^{2}+(y-(-3))^{2}=5^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "y=-\\sqrt{5^{2}-(x-2)^{2}}-3",
"displayFormula": "(x-2)^{2}+(y-(-3))^{2}=5^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
}
]
},
"pointsToDraw": {
"pointsLatex": [
"(2,-3)"
],
"pointsDecimal": [
{
"fst": 2,
"snd": -3
}
],
"attributes": [
{
"color": "PURPLE",
"labels": [
"\\mathrm{Center}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
}
]
},
"linesToDraw": [
{
"p1x": "2",
"p1y": "-3",
"p2x": "5.5355339059327",
"p2y": "0.5355339059327",
"attributes": {
"color": "GRAY",
"lineType": "BOLD",
"labels": [
"\\mathrm{radius=}5"
],
"isAsymptote": false
}
}
],
"functionChanges": [
{
"origFormulaLatex": [],
"finalFormulaLatex": [],
"plotTitle": "(x-2)^{2}+(y-(-3))^{2}=5^{2}",
"paramsLatex": [],
"paramsReplacementsLatex": []
}
],
"localBoundingBox": {
"xMin": -9.25,
"xMax": 13.25,
"yMin": -14.25,
"yMax": 8.25
}
},
"showViewLarger": true
}
}
}
Solution
Solution
Solution steps
Rewrite in the form of the standard circle equation
Therefore the circle properties are:
Graph
Popular Examples
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Frequently Asked Questions (FAQ)
What is x^2-4x+y^2+6y-12=0 ?
The solution to x^2-4x+y^2+6y-12=0 is Circle with (a,b)=(2,-3),r=5