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.

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

idk

step-by-step explanation: