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The general form of the equation of a circle is 7x2 + 7y2 − 28x + 42y − 35 = 0. The equation of this circle in standard form is ____. The center of the circle is at the point____ , and its radius is_____ units.

Respuesta :

calculista
7x2 + 7y2 − 28x + 42y − 35 = 0
7/7x2 + 7/7y2 − 28/7x + 42/7y − 35/7 = 0

x² + y² - 4 x + 6 y - 5 = 0
( x² - 4 x + 4 ) + ( y² + 6 y + 9 ) - 4  - 9 - 5 = 0
The equation in the standard form is:( x - 2 )² + ( y + 3 )² = 18
                                                           ( x - 2 )² + ( y - (-3) )² = 18

The center is at the point ( 2, - 3 ).Its radius is: √18 = 3√2 units

 the answer is 
a) (x-2)²+(y+3)²=18
b) The center of the circle is at the point (2,-3)
c) the radius is 3√2 units
frika

The equation of a circle is [tex]7x^2 + 7y^2 - 28x + 42y - 35 = 0.[/tex]

First divide the equation by 7 and group terms with x and with y:

[tex]x^2 -4x+ y^2 + 6y - 5 = 0,\\ \\(x^2-4x)+(y^2+6y)-5=0.[/tex]

In both brackets write a perfect square:

[tex](x^2-4x+4-4)+(y^2+6y+9-9)-5=0,\\ \\(x-2)^2-4+(y+3)^2-9-5=0,\\ \\(x-2)^2+(y+3)^2=18.[/tex]

The equation of this circle in standard form is [tex](x-2)^2+(y-(-3))^2=18.[/tex] The center of the circle is at point (2,-3) and the radius is [tex]\sqrt{18}=3\sqrt{2}[/tex] units.

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