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The equation of a circle in general form is ​x2+y2+20x+12y+15=0​ .




What is the equation of the circle in standard form?
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Sagot :

sreedevi102

Answer:

[tex](x +10)^2 + (y+6)^2 = 121[/tex]

Step-by-step explanation:

Given equation :

                        [tex]x^2 + y^2 + 20x +12y + 15 = 0[/tex]

Standard equation of circle :

                                            [tex](x - a)^2 + (y-b)^2 = r^2[/tex]

We will consider the x terms in the equation first to find a.

[tex](x-a)^2 = x^2 - 2ax + a^2[/tex]

-2ax = 20x

a = -20x/2x = -10

a = -10

Next consider the y terms to find b.

[tex](y - b)^2 = y^2 -2by + b^2[/tex]

-2by = 12y

b = -12y /2y = -6

b = -6

[tex](x+10)^2 + (y-6)^2 = x^2 +20x +100 + y^2 +12y + 36[/tex]

                             [tex]=x^2 + y^2 + 20x +12y + 136\\[/tex]

But the given equation constant is 15. So find the difference between 136 and 15 to find the radius : 136 - 15 = 121

Therefore, radius = √121 = 11

Equation in standard form :

                                         [tex](x +10)^2 + (y+6)^2 = 121[/tex]

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