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Sagot :
Find the centre first by reoriganisation it into the form (x - cx)^2 + (y -cy)^2 = r^2
x^2 + y^2 + 6x -6y +5 = 0
(x + 3)^2 - 9 + (y - 3)^2 - 9 + 5 = 0
(x + 3)^2 + (y - 3)^2 = 13
The radius does not matter the centre is ( -3, +3)
The diameter must pass through the center so substituting into the equation for the line
2x-y+a = 0
-6 -3 + a = 0 so a = 9
The value of 'a' in the equation (2x - y + a = 0) is 0 and this can be determined by using the generalized equation of a circle.
Given :
Circle Equation -- [tex]x^2-2x+y^2-4y-4[/tex]
Line equation -- 2x - y + a = 0
First, convert the given equation of a circle in the generalized equation of a circle which is given by:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where (a,b) represents the center of the circle and 'r' is the radius of the circle.
The generalized form of the given equation of a circle is:
[tex](x-1)^2-1+(y-2)^2-4-4=0[/tex]
[tex](x-1)^2+(y-2)^2=3^2[/tex]
So, the center of the circle is (1,2) and the radius is 3.
The center of the circle satisfy the line passing through it, that is:
2(1) - (2) + a = 0
a = 0
So, the value of a in the equation (2x - y + a = 0) is 0.
For more information, refer to the link given below:
https://brainly.com/question/23799314
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