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Which the points (5,7) and (6,-2) lie on the locus represented by the equation xsquare+ysquare-4x-6y=12

Sagot :

Answer:

(5, 7)

Step-by-step explanation:

What is a Locus

Locus is the set of all points that satisfy a mathematical condition(s) that forms into a line/curve, or in 3-D, a surface.

In our case, all points equidistant from another point by an x distance is a circle, where x is the radius of the circle.

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Solving the Problem

Geometric Solution

We're given the equation of a circle--by noticing the terms [tex]x^2,\: y^2[/tex] and their equivalent coefficient of 1--, and two points that could satisfy this circle's condition.

We need to find the standard form of the given equation to identify its radius and center, and then calculate the distance from its center to each point!

(The standard form of a circle is:[tex](x-h)^2+(y-k)^2=r^2[/tex])


We can complete the square to find the circle's equation,

                                [tex]x^2-4x+y^2-6y=12[/tex]

                    [tex]x^2-4x+4+y^2-6y+9=12+4+9[/tex]

                               [tex](x-2)^2+(y-3)^2=25[/tex]

The circle's center is at (2,3) and has a radius of 5.

Using the distance formula we can find the distance between each point and the center; whichever has a distance of 5 (the condition) is a part of the locus.

Distance between (2,3) and (5,7):

                        [tex]\sqrt{(5-2)^2+(7-3)^2} =\sqrt{3^2+4^2} =5[/tex].

Distance between (2,3) and (6,-2):

                     [tex]\sqrt{(6-2)^2+(-2+3)^2}=\sqrt{4^2+1^2}=\sqrt{17} =4.12[/tex].

So, (5,7) is our answer!

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Algebraic Solution

If a point satisfies the conditions set by a mathematical function/equation, then plugging the two points into the given circle equation and determining if the left and right sides agree with each other can get us our final answer!

Plugging (5,7):

                                   [tex]5^2+7^2-4(5)-6(7)=12[/tex]

                                      [tex]25+49-20-42=12[/tex]

                                                     [tex]12=12[/tex].

Plugging (6,-2):

                              [tex]6^2+(-2)^2-4(6)-6(-2)=12[/tex]

                                     [tex]36+4-24-12=12[/tex]

                                                  [tex]4\neq 12[/tex].

Only (5,7) satisfy the equation, so that must be a part of the circle's graph or its locus!

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