Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
