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

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

oof

step-by-step explanation: