Answer:
Step-by-step explanation:
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' is 0 and this can be determined by using the standard form of the equation of a circle.
Given :
Equation of a Circle -- [tex]x^2-2x+y^2-4y-4=0[/tex]
Equation of a Line -- 2x - y + a = 0
The following steps can be used in order to determine the value of 'a':
Step 1 - First make the equation of a circle in standard form.
[tex]x^2-2x+y^2-4y-4=0[/tex]
[tex](x-1)^2-1+(y-2)^2-4-4=0[/tex]
[tex](x-1)^2+(y-2)^2 = 3^2[/tex]
Step 2 - From the above step, it can be concluded that the center of the circle is (1,2).
Step 3 - If (1,2) is the center of the circle then it also satisfies the equation of diameter (2x - y + a = 0).
2x - y + a = 0
2(1) - 2 + a = 0
a = 0
For more information, refer to the link given below:
https://brainly.com/question/23528207
