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Consider the equation of a circle x^2-2x+y^2-4y-4. If the line 2x-y+a=0 is its diameter. Then find the value of a.

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